TABLE OF CONTENTS CHAPTER 1 LIMIT AND CONTINUITY... 26

TABLE OF CONTENTS CHAPTER LIMIT AND CONTINUITY... LECTURE 0- BASIC ALGEBRAIC EXPRESSIONS AND SOLVING EQUATIONS... LECTURE 0- INTRODUCTION TO FUNCTIONS... 9 LECTURE 0- EXPONENTIAL AND LOGARITHMIC FUNCTIONS... 7 LECTURE 0- TRIGONOMETRIC FUNCTIONS... CHAPTER LIMIT AND CONTINUITY... 6 LECTURE - INTRODUCTION TO LIMITS... 6 LECTURE - TECHNIQUES FOR COMPUTING LIMITS... 9 LECTURE - LIMITS AT INFINITE... LECTURE - PIECEWISE CASE AND INFINITE LIMITS... 6 LECTURE - SQUEEZE THEOREM AND TRIGONOMETRIC LIMITS... 7 LECTURE -6 CONTINUITY... 80 LECTURE -7 DEFINITION OF THE LIMIT... 89 PRACTICE FOR CHAPTER... 9 CHAPTER DERIVATIVES... 97 LECTURE - DEFINITION OF THE DERIVATIVE... 97 LECTURE - BASIC DERIVATIVES... 09 LECTURE - PRODUCT AND QUOTIENT RULES... LECTURE - CHAIN RULE... LECTURE - IMPLICIT DERIVATIVES... 7 LECTURE -6 DERIVATIVE OF INVERSE/HYPERBOLIC FUNCTIONS... LECTURE -7 THE TANGENT LINE TO A GRAPH OF A FUNCTION... 6 LECTURE -8 L`HOPITAL S THEOREM... 7 LECTURE -9 NEWTON S METHOD AND LINEAR APPROXIMATIONS... 78 PRACTICE FOR CHAPTER... 86 CHAPTER APPLICATION OF DERIVATIVES... 89 LECTURE - LOCAL EXTREMA ON AN INTERVAL... 89 LECTURE - CONCAVITY... LECTURE - SUMMARY OF CURVE SKETCHING... LECTURE - INTERMEDIATE VALE THEOREM AND MEAN VALUE THEOREM... 8 LECTURE - OPTIMIZATION PROBLEMS... 6 LECTURE -6 RELATED RATES... 6 PRACTICE FOR CHAPTER... 6 CHAPTER INTERGRALS... 68 LECTURE - THE ANTI- DERIVATIVES... 68 LECTURE - BASIC INTEGRALS... 78 LECTURE - THE SUBSTITUTION RULE IN INTEGRATIONS... 8 LECTURE - MORE SUBSTITUTION RULE IN INTEGRATIONS... 97 LECTURE - INTEGRATION BY PARTS... 09 LECTURE -6 RIEMANN SUMS... 7 LECTURE -7 RIEMANN INTEGRALS... 0 LECTURE -8 FUNDAMENTAL CALCULUS THEOREM... 7 LECTURE -9 MEAN VALUE THEOREM OF INTEGRAL... 9 LECTURE -0 THE AREA OF A REGION BETWEEN TWO CURVES... PRACTICE FOR CHAPTER... 6

CHAPTER 0. REVIEW LECTURE 0- BASIC ALGEBRAIC EXPRESSIONS AND SOLVEING EQUATIONS FACTORING: Factoring is the process of finding the factors (finding what to multiply together to get an epression). It is like "splitting" an epression into a multiplication of simpler epressions. ) Greatest common factor (GCF): The GCF is the largest factor two or more epressions have in common. ) Grouping (usually four terms case): Factoring by grouping means that you will group terms with common factor before factoring. ) Trinomial factor: ac + (ad + bc) + bd = (a + b)(c + d) ) A B = (A B)(A + B) ) A + B : Prime factor 6) A B = (A B)(A + AB + B ) 7) A + B = (A + B)(A AB + B ) 8) Substitution factoring Eample : Factor completely the epression: y 0 y y We can find GCF y. y 0 y y = y (y y ) Eample : Factor completely the epression: a b a + b. This consists of terms and we are going to use Grouping a b a + b = (a b) (a b) = (a b)( ) Eample : Factor completely the epression: 9 + Find two integers a and b such that its sum is 9 and the product of two is () = 8 a b 8 8 Then 9 + = 8 + = ( ) ( ) by grouping = ( )( ) by GCF

Eample : Factor completely (A) 8 7a (B) + (A) 8 7a A B case = () (a) Let A = and B = a = ( a)[() + a + (a) ] = ( a)( + 6a + 9a ) (B) + By grouping = ( ) ( ) By GCF = ( )( ) A B = (A B)(A + B) where A = and B = = ( )( )( + ) Eample : Factor completely the epression: (A) 0 + 9 (B) e e 6 We use the substitution method and trinomial factor method. (A) 0 + 9 = ( ) 0( ) + 9 A 0A + 9 = (A )(A 9) where A = = ( )( 9) A B cases = ( )( + )( )( + ) (B) e e 6 = (e ) (e ) 6 A A 6 = (A )(A + ) where A = e = (e )(e + ) Eample 6: Factor completely the epression: (A) 8 (B) + + 0 (C) 7 (D) 6 + + (E) 9 (F) 9y (G) + (H) + (I) 8 (J) + 6 (K) 6 (L) sin (M) (N) cos cos 8 (A) ( )( + ) (B) ( + )( + 6) (C) ( + )( ) (D) ( + )( + ) (E) ( )( + ) (F) ( y)( + y) (G) ( )( + )( + ) (H) Prime (I) ( )( + + ) (J) ( + )( + 6) (K) ( )( + ) (L) ( sin )( sin + ) (M) ( )( + ) (N) (cos )(cos + )

SOLVING A EQUATION: A solution is a value we can put in place of a variable (such as ) that makes the equation true ) When it is a linear equation a + b = 0, = b a ) When it is a quadratic equation a + b + c = 0 (a 0), = b ± b ac a ) When it is a higher degree polynomial, we can use substitution or Grouping. Eample 7: Find all real or comple solutions of the following equation: (A) 6 = (B) + = 0 (C) + + 0 = 0 (A) 6 = 0 Make one side zero = ( 6) ± ( 6) ()( ) () = 6 ± 8 = ± By Quadratic formula (B) + = 0 = ( ) ± ( ) ()() () Make one side zero By Quadratic formula = 8 = (C) + = 0 + () + = 0 + ( + ) = 6 We use the perfect square form Add square of (half of the coefficient of ) to both sides Simplify + = ± 6 = ± 6 i Eample 8: Find all real or comple solutions of the following equation. (A) 8 = (B) ( ) + 7 = 0 (A) ± (B) ± i

Eample 9: Find all real or comple solutions of the following equation: (A) 8 = 0 (B) ( ) ( ) = 8 (C) 8 9 = 0 (A) 8 = 0 ( 8) = 0 ( )( + ) = 0 First, find the greatest common factor Factor the trinomial Therefore, =, 0, or (B) ( ) ( ) = 8 ( ) ( ) 8 = 0 ( )( + ) = 0 Make one side zero Use substitution method and it is of the form A A 8 ( 7)( ) = 0 Therefore, = 7, (C) 8 9 = 0 ( ) 8( ) 9 = 0 Use substitution method and it is of the form A 8A 9 ( 9)( + ) = 0 ( )( + )( + ) = 0 Factor more =,, or = =,, ±i Eample 0: Find all real or comple solutions of the following equation. (A) = (B) = (C) 9 = 0 (D) + = 0 (E) ( + ) 7( + ) + = 0 (F) ( + ) ( + ) 8 = 0 (G) 8 + = 0 (H) + = 0 (I) 0 + 9 = 0 (J) 6 = 0 (A) 0, (B) 0,7 (C) 0,, (D), 0, (E), (F) (G) (H), 8, ±, ±i (I) ±, ± (J) ±, ±i

6 PRINCIPAL n-th ROOT n a = b means that a = b n ; a is the radicand and n is the inde. NOTE: When a, b 0 n ab n a b n = a n = a n b n b n a n a, if n is even = { a, if n is odd NOTE: For any real number a, n a m n m = a if it is defined m n a mn = a if it is defined Eample : Simplify the epression: (A) ( 7) (B) ( ), > (C) ( + ), < (D) +, < 0 Solution (A) ( 7) = 7 = 7 (B) ( ), > =, > 0 is positive since > = (C) ( + ), < = +, + < 0 + is negative since < = ( + ) = (D) +, < 0 = + < 0 and > 0 since < 0 = + ( ) = Eample : Simplify the epression: (A) ( 6) (B) ( + ), > (C) ( ), < (D), > 0 (A) 6 (B) + (C) + (D)

7 Eample : Rewrite the epression using eponential form. Assume that it is defined. (A) (A) = ( ) = (B) + y (C) (D) (B) + y (C) = / = / (D) = = = ( + y ) Warning ( + y ) + y Eample : Rewrite the epression using eponential form. Assume that it is defined. 6 (A) 8 (B) y (C) (D) (A) (B) ( y ) (C) (D) Eample : Rationalize the numerator and simplify: ( )( + ) = ( )( + ) = ( )( + ) = + Multiple conjugate of numerator in both numerator and denominator Simplify numerator and simplify Eample 6: Rationalize the numerator and simplify (A) + (B) ++ (A) (B) ++ + Eample 7: Rationalize the denominator and simplify (A) 9 (B) + + (A) + (B) + +

8 Eample 8: Solve the equation: + = + = + ( + ) = ( + ) Add to both sides Take square to both sides + = + 6 + 9 0 = + + Simplify 0 = ( + )( + ) Factor Then =, are possible solutions. Check your answer. When =, + = is true When =, + is false Therefore, is only solution of this equation. Eample 9: Solve the equation: 6 = 6 = ( 6) = Add to both sides Take square to both sides 6 = 0 = + 6 Simplify 0 = ( )( ) Factor Then =, are possible solutions. Check your answer. When =, () 6 = is true When =, () 6 = is true Therefore, the set of solutions is {,}. Eample 0: Solve the equation (A) + = (C) (B) 6 = = (D) = (A) (B), (C) (D)

9 LECTURE 0- INTRODUCTION TO FUNCTIONS RELATION: A relation is just a set of ordered pairs. FUNCTIONS: A function is a relation between a set of inputs (domain) and a set of permissible outputs (codomain) with the property that each input is related to eactly one output. The domain (input) is the set of all input values to which the rule applies. These are called your independent variables. The range (output) is the set of all output values. These are called your dependent variables VERTICAL LINE TEST: A graph is the graph of a function if and only if there is no vertical line that crosses the graph more than one. FUNCTION FUNCTION NOT A FUNCTION Fail the vertical line test pass the vertical line test pass the vertical line test Eample : Determine whether the relation represents y as a function of. (A) y = + (A) By the vertical line test, it is a function (B) + y = (B) Since y = y = ±, (0,) and (0, ) are points on the graph of the curve. An input = 0 has two different output y-values (NOT only one y-value). Therefore, it is Not a function. Eample : Determine whether the relation represents y as a function of. (A) y = + (B) y = (C) y = + (D) y = + (E) y = + (F) y = + (A) Function (B) Function (C) Not a function (D) Function (E) Not a function (F) Function

0 Eample : Consider f() = + 7. Find the following. (A) f( ) (B) f() (C) f( + a) (D) f(+a) f() a (A) f( ) Put ( ) in the place of = ( ) ( ) + 7 = (B) f() = () () + 7 Put () in the place of = (9 ) 9 + 7 () = 9 by FOIL = 8 9 + 7 Simplify (C) f( + a) = ( + a) ( + a) + 7 Put ( + a) in the place of = ( + a + a ) a + 7 ( + a) = ( + a)( + a) by FOIL = + a + a a + 7 Simplify (D) Use the result of (B) f( + a) f() a = ( + a + a a + 7) ( + 7) a = a + a a a a( + a ) = a Simplify Factor out a in the numerator. = + a Simplify Eample : Consider f() = +. Find the following. (A) f() (B) f( ) (C) f() (D) f( + ) (A) (B) 8 (C) 7 6 + (D) + + 6 (E) + (E) f( ) (G) f( + a) (F) f() (H) f(+a) f() a (F) 9 6 + (G) + 6a + a a (H) 6 + a

DEFINITION: ) The function y = f() is an even function if it is symmetry with respect to y-ais: f( ) = f() ) The function y = f() is an even function if it is symmetry with respect to the origin: f( ) = f() EVEN FUNCTION ODD FUNCTION Graph y ais ( a, b) (a, b) y ais (a, b) ais ais ( a, b) Symmetry about the y-ais Symmetry about the origin Algebraically test f( ) = f() f( ) = f() Eample : Determine each function is even, odd, or neither. (A) f() = (B) f() = (C) f() = (A) f( ) = ( ) = = f() Then, it is an even function. (B) f( ) = ( ) ( ) = + = ( ) = f() Then it is an odd function. (C) f( ) = ( ) ( ) = +. Then, f( ) f() and f( ) f(). It is neither even nor odd. Eample 6: Determine each function is even, odd, or neither. (A) f() = + 7 (B) f() = + (C) f() = 9 (E) f() = ln( ) (D) f() = (F) f() = (A) Even (B) Neither (C) Odd (D) Odd (E) Neither (F) Even

HOW TO FIND DOMAIN OF A FUNCTION: CASE DOMAIN Polynomials/ Odd inde radicals Rationales Even inde radicals Logarithmic case All real numbers All real numbers ecept the -values which make denominator zero All real numbers which make the inside of radical non-negative All real numbers which make the inside of logarithm positive. Eample 7: Find the domain of the following functions Solution (A) y = + (B) y = (C) y = (D) y = + (A) Since it is a polynomial, the domain is (, ). ( )(+), < (E) y = log ( ) + 7 (F) y = {, (B) Since it is a square root function, the y is defined for 0 The domain is (, ] in the interval notation. (C) Since it is a rational function, the denominator cannot be zero: and The domain is (, ) (,) (, ) in the interval notation. (D) Since it is a rational function which contains a radical, it satisfies the following two conditions: Radical condition Rational condition + 0 Then, and.the domain is [,) (, ) in the interval notation. (E) Since it is a logarithmic function, the inside of the logarithm is positive: > 0 > The domain is (, ) in the interval notation. (F) The y is defined for < or. The domain is (, ) [, ) = (, ) in the interval notation. Eample 8: Find the domain of the following functions (A) y = + (B) y = (C) y = + (E) y =, < < 0 (G) y = {, 0 (D) y = (F) y = 6 (H) y = + (I) y = ln( + 7) (J) y = log( ) (A) (, ) (B) (, ) (C) (, ) (, ) (D) (, ) (,) (, ) (E) [, ) (F) (, ] (G) (, ) (H) [,0) (0, ) (I) (7, ) (J) (,)

INCREASING/ DECREASING/ CONSTANT A function f is increasing on an interval if for any a and b in this interval, a < b implies f(a) < f(b) A function f is decreasing on an interval if for any a and b in this interval, a < b implies f(a) > f(b) Find INCREASING/DECREASING/CONSTANT INTERVAL : intervals and open interval. decreasing y y = f() increasing increasing decreasing LOCAL EXTREMA We say that f() has a local minimum at = a if f(a) f() for every in some open interval around = c. decreasing y ais Local ma y = f() increasing We say that f() has a local maimum at = a if f(a) f() for every in some open interval around = c. ais increasing decreasing Local min Local min Eample 9: Find the increasing/decreasing/constant intervals of the graph of the function. (A) (B) (A) II: (, ) (, ) DI: (, ) (, ) No constant (B) II: (, ) DI: (, ) CI:(,) Eample 0: Find the local maimum and local minimum of a function in the figure if any (A) (B) (A) (B) Local ma at = and Local min 0 at = Local ma at = and Local min at = 0

DEFINITION: Let f and g be functions. The composite function, denoted by f g (read as f composed with g ), is defined by (f g)() = f(g()) The domain of the composite function of f g is the set of all such that is in the domain of g and g() is in the domain of f. Eample : Let f() = and g() = +. Find (A) (g f)() (B) (f g)( ) (A) (g f)() = g(f()) = g( ) = ( ) ( ) + = (9 6 + ) + + FOIL = 8 8 + 0 Simplify (B) (f g)( ) = f(g( )) = f() g( ) = ( ) ( ) + = 8 + 0 + = = () = 6 Eample : Let f() = and g() = + 6. Find each of the composite functions. (A) (f g)() (B) (g f)( ) (C) (f g)() (D) (g f)() (A) (B) (C) 6 9 + 8 (D) 8 9 + 6 Eample : Let f() = + and g() = +. Find each of the composite functions. (A) (f g)() (B) (g f)() (A) + + (B) + 7 + 6 Eample : Let f() = and g() =. Find (f g)() and its domain. (f + g)() = + + Domain: {, }

ONE TO ONE FUNCTIONS One to one function (Invertible function): A function for which every element of the range of the function corresponds to eactly one element of the domain. HORIZONTAL LINE TEST (One to one function or not): A test uses to determine if a function is one-to-one. If a horizontal line intersects a function's graph more than once, then the function is not one-to-one. Eample : Use the graph to decide whether it is one to one function or not. (A) y = + (B) y = + (C) y = + (D) y = (A) Not (B) One to one (C) One to one (D) Not DEFINITION: Let f and g be functions such that f(g()) = for every in the domain of g, g(f()) = for every in the domain of f. ) The function g is the inverse of the function f. It is denoted by g = f. (Warning: f ) f ) The domain of f is equal to the range of f, and vice versa. (,) (0,) y = f() (,) y = f () (,) (,0) (, ) Eample 6: Find the inverse function of the function f if any: f() =. We know that it is one to one by the horizontal line test. So, it has its inverse. 6 The inverse function of y = is = y y = + Put y in the place of f() Inverse function: echange y Solve for y y = + f () = +

6 Eample 7: Find the inverse function of the function f if any: f() = ( ),. We know that it is one to one by the horizontal line test on the interval (, ]. So, it has its inverse. The inverse of y = ( ), is Replace f() by y = (y ), y Inverse function: interchange y + = (y ), y y = ± +, y y = +, Solve for y But y Add in both sides y = + Therefore, f () = + Replace y by f () Eample 8: Find the inverse of the function f(). (A) f() = (B) f() = + (C) f() = ( ) 7 + 7 (D) f() = + 8 (E) f() = (G) f() =, 0 (I) f() = + (F) f() = 8+7 (H) f() = ( + ), (J) f() = (A) f () = + (B) f () = (C) f () = 7 + (D) f () = ( 8) 7 + (E) f () = (F) f () = 7 8 (G) f () = + (H) f () = (I) f () =, 0 (J) f () = ( ),

7 LECTURE 0- EXPONENTIAL AND LOGARITHMIC FUNCTIONS EXPONENTIAL FUNCTION: An eponential function f with base a is a function of the form f() = a eponent, or y = a base, where a is real numbers such that a > 0 and a is a real number. ) Domain: (, ) ) Range: (0, ) since a > 0 ) f is a one to one, continuous and smooth function (no sharp points) ) a > 0 < a < y = a, a > y = a, 0 < a < As +, f() + and As, f() +0 As +, f() +0 and As, f() + LOGARITHMIC FUNCTION: For a > 0 and a, the logarithmic function with base a is denoted f() = log a (), where y = log a () if and only if a y = (it is the inverse function of y = a ) ) When a = 0, we write log 0 () = log() ; the common logarithm. ) When a = e, we write log e () = ln() ; the natural logarithm. ) Domain of y = log a () is the interval (0, ). ) Range of y = log a () is the interval (, ). ) Vertical Asymptote: = 0. 6) a > 0 < a < y = log a(), a > y = log a(), 0 < a < As +0, f() and As +, f() + As +0, f() + and As +, f()

8 PROPERTIES OF LOGARITHM: log a A where a > 0 and a. ) When a = 0, log 0 (A) = log(a) ) When a = e, log e (A) = ln(a) ) log a A = log A, > 0 and ) log a a log a(t) = T ) log a () = 0 6) log a (a) = 7) log a (AB) = log a (A) + log a (B) 8) log a ( A B ) = log a (A B ) = log a(a) log a (B) 9) log a (A n ) = n log a (A) Eample : Evaluate the epression by using the properties of logarithm (A) log 7 (9) (B) log () (A) log 7 (9) Use ) to change the base. = ln(9) ln(7) = ln( ) ln( ) = ln() ln() By 9) Simplify = (B) log () By ) = log ( ) = = 9 Eample : Evaluate the epression by using the properties of logarithm. (A) log 7 () (B) log () (C) log () (D) log 8 () (E) log / ( ) 7 (F) log /(8) (G) log () (H) e ln() (A) 0 (B) (C) (D) (E) (F) (G) 7 (H)

9 Eample : Epand the logarithmic epression: log ( y ) where all variables are positive log ( y w ) = log ( y w ) = log( y ) log(w ) By 8) = log( ) + log(y ) log(w ) By 7) = log() + log(y) log(w) By 9) w Epress it as eponent forms Eample : Make the following epression as one logarithmic form: log log y log w Solution log log y log w = log ( ) log (y ) log (w ) By 9) = log ( = log ( y ) log (w ) By 8) y w ) By 8) = log ( y w ) Simplify = log ( y w ) Eample : Epand the logarithmic epression: (A) ln( y + ) (B) log ( (+) ( ) (+) 7 ) (C) log( + ) (D) ln ( ( ) (+) 7) Eample 6: Make the following epressions as one logarithmic form. (A) ln ln y ln z (A) ln + ln(y + ) (B) log( + ) (C) log( + ) + (D) (A) log( ) 7 log( + ) ln ln( ) 7 ln( + ) ln ( y z ) (B) log ( ( ) ( ) ) (B) log( ) log() + log( )

0 Eample 7: Write each equation in its equivalent eponent/logarithmic form. (A) log() = a (B) = a (A) First find the base: 0 log() = a log 0 () = a = 0 = 0 a : First, find the base (B) First find the base: = a = log ( ) = log (a) : First, find the base Eample 8: Write each equation in its equivalent eponent form. (A) log () = (B) log a () = (A) = (B) = a Eample 9: Write each equation in its equivalent logarithmic form. (A) = (B) = (A) = log (B) = log Eample 0: Solve the equation: = e + = e + ln() = ln(e + ) ln() = ( + ) ln(e) Take the natural logarithm of each side Since ln e = ln() = + = + ln() Eample : Solve the equation: ln( ) = ln( ) = e ln( ) = e = e = e + Eponentiate each side Since a log a T = T Simplify

Eample : Solve the equation: ln() + ln( ) = ln() ln() + ln( ) = ln() ln[( )] = ln() = Using log property, make one log in left side The arguments are equal since logarithmic function is one to one = 0 ( )( + ) = 0 =, Check your answers: Left side Right side ln() + ln( ). 89 log (). 0 ln( ) + ln( ) is undefined log (). 0 The solution set is {} since Eample : Solve the equation: log () + log ( ) = log () + log ( ) = log (( )) = = Using log property, make one log in left side The arguments are equal since logarithmic function is one to one = 8 8 = 0 ( )( + ) = 0 =, Check your answers: Left side Right side log + log = log ( ) + log ( ) is undefined The solution set is {} since Eample : Solve the equation (A) e = (B) ln( + ) = (C) log () + log ( + ) = log () (D) log ( + ) log () = log () (E) log () log () = log ( + ) (F) log( ) + log() = log( + ) (A) (B) e (C) (D) (E) No solution (F)

LECTURE 0- TRIGONOMETRIC FUNCTIONS TRIGONOMETRY: Let θ be a real number and let P = (, y) be the point in the plane. Let r = + y : radius of the circle. sin θ = y r csc θ = sin t = r y-ais cos θ = r sec θ = cos t = r sin t tan θ = cos t = y cot θ = tan t = y -ais r θ (, y) Eample : Solve the equation on the interval [0,π): cos + = 0. cos + = 0 cos = Subtract from both sides cos = Divide both side by In the interval [0,π), there are two angles such that cos = : = π and = π. The set of solutions is { π, π }. Eample : Find the eact value of (A) cos ( π ) (B) csc (π ) (C) sec ( π 6 ) (D) tan (π ) (A) (B) (C) (D) Eample : Solve the equation on the interval [0,π): sin = 7π, π 6 6 Eample : Solve the equation on the interval [0,π): tan = π, π Eample : Solve the equation on the interval [0,π): cos ( π ) = 0, π, π, π

TRIGONOMETRIC FORMULAS: Fundamental Identities Sum and difference Formulas tan θ = sin θ cos θ cot θ = cos θ sin θ cot θ = tan θ csc θ = sin θ sec θ = cos θ sin θ + cos θ = tan θ + = sec θ cot θ + = csc θ sin(a + b) = sin a cos b + cos a sin b sin(a b) = sin a cos b cos a sin b cos(a + b) = cos a cos b sin a sin b cos(a b) = cos a cos b + sin a sin b tan(a + b) = tan(a b) = tan a + tan b tan a tan b tan a tan b + tan a tan b Half angle Formulas Double angle Formulas cos ( θ + cos θ ) = sin(θ) = sin θ cos θ sin ( θ cos θ ) = tan(θ) = tan θ tan θ tan ( θ cos θ ) = sin θ cos(θ) = cos θ sin θ cos(θ) = cos θ cos(θ) = sin θ Eample 6: Find the eact value of tan( ) ) By sum and difference formula: tan( ) = tan( 0 ) = tan( ) tan(0 ) + tan( ) tan(0 ) = + = 6 = = + 6 ) By the half-angle formula tan( ) = tan ( 0 ) = cos(0 ) sin(0 ) = = = Eample 7: Find the eact value of cos(θ) if sin θ =. cos(θ) = sin θ Since we only know sin θ = = ( ) = 7

Eample 8: Find the eact value of (A) 6+ (A) sin(0 ) (B) cos(7 ) (B) 6 Eample 9: Write the following in a single trigonometry function. (A) sin() cos() + cos() sin() (B) cos() cos() + sin() sin() (A) sin(8) (B) cos() Eample 0: Evaluate of cos() if (A) sin() = (B) cos() = (A) 7 (B) 8 Eample : Evaluate sin( ) by half angle formula. Eample : Prove the identity by using the trigonometric formula: cos sin cos = sin = = cos ( cos )( + cos ) + cos cos sin = + cos sin + cos = sin = cos = ( cos )( + cos ) Eample : Prove the identity by using the trigonometric formula: (sin +cos ) + sin cos = Eample : Prove the identity by using the trigonometric formula: sin cos = sec tan s Eample : Prove the identity by using the trigonometric formula: cos sin = +sin cos

INVERSE TRIGONOMETRIC FUNCTIONS The function Inverse function Domain of inverse Range of inverse y = sin y = sin or y = arcsin [,] [ π, π ] y = cos y = cos or y = arccos [,] [0, π] y = tan y = tan or y = arctan (, ) ( π, π ) y = csc y = csc or y = arccsc (, ] [, ) [ π, π ], y 0 y = sec y = sec or y = arcsec (, ] [, ) [0, π], y π y = cot y = cot or y = arccot (, ) (0, π) Eample 6: Evaluate (A) sin ( ) (B) cos (arcsin ( )) (C) arcsin (cos (π 6 )) (A) sin ( ) Find angle y such that sin y = and π y π = π (B) cos (arcsin ( )) sin ( π ) = and π is in [ π, π ] Let y be an angle such that sin y =, π y π = cos y A point on the terminal side of angle y is (, ). = (C) arcsin (cos ( π 6 )) cos ( π 6 ) = = arcsin ( = π ) Find angle y such that sin y = and π y π Eample 7: Evaluate (A) tan ( ) (B) sin ( ) (C) arccos ( ) (D) cot ( ) (E) sin (arccos ( )) (F) cos (arctan ( )) (G) tan (arccos ( )) (H) cos (arcsin ( )) (A) (B) (C) (D) (E) (F) π π 6 π π (G) (H)

6 CHAPTER. LIMITS AND CONTINUITY LECTURE - LIMITS: GRAPHICAL AND NUMERICAL APPROACH DEFINITION: Assume that y = f() is defined in some open interval containing a. Finding a it entails understanding how a function behaves near a particular value of. ) If f() approaches L for all sufficiently close to a and < a without letting = a, then we said that the left it of f() is L as approaches a and write this as f() = L. a ) If f() approaches L for all sufficiently close to a and > a without letting = a, then we said that the right it of f() is L as approaches a and write this as f() = L. a + ) If f() approaches L as approaches a without letting = a, then we said that the it of f() is L as approaches a and write this as a f() = L. In other words, if f(), f() eist and f() = f() = L, then a + a + a a a f() = L. Eample : Find f() by using the graph of f() = +. We can use the graph of f or the value table to find the it. f() = + The graph of f() indicates that f() is close to as is close to from both sides. Therefore, f() = f() = +.9. 9.99. 99.999. 999.00. 00.0. 0.. For numerical view, we make the value table to decide whether f() is close to a number as is close to from both sides. The value of f() is close to as is close to from both sides. In that case, we say that f() =

7 Eample : Use the graph of f() as shown in the below figure to find f() + f() f() (D) f(). The it of f() at = a does not depend on f(a). (A) The graph of f() indicates that f() is close to as is close to from the left side. Therefore, f() = (B) The graph of f() indicates that f() is close to as is close to from the right side. Therefore, f() = + (C) The graph of f() indicates that f() is close to as is close to from both sides. Therefore, f() = (D) The value of f() is in the graph. f() = f() = (It is possible that f() f())

8 Eample : Use the graph of f() = f() + f() f() (D) f(). in the right figure to find The it of f() at = a does not depend on how the function is defined at the point = a. (A) As is close to from the left side, f() is close to : f() = (B) As is close to from the right side, f() is close to : f() = + (C) We know that f() is undefined. But as is close to from both sides, f() is close to : f() = f() + Then f() = (D) The f() is undefined since it has an open point at =. Even if f is undefined at =, the it of f at = does eist

9 Eample : Use the graph of f() in the right figure to find f() + f() f() (D) f(). The it of f() at = a does not depend on how the function is defined at the point = a. (A) As is close to from the left side, f() is close to : f() = (B) As is close to from the right side, f() is close to : f() = + (C) As is close to from both sides, f() is close to different values from each side: Then f() f() + f() does not eist. (D) The function f is defined at = (it has a closed point at = ): (,) f() =

0 Eample : Use the graph of f() to find f() f() + (A) (B) (C) (D) y = f() f() (D) f( ) (E) (F) (E) f() (F) f() + (G) No it (H) (G) f() (H) f() Eample 6: Use the graph of f() to find y = f() 0 f() 0 f() f() 0 + (D) f(0) (A) (B) 0 (C) No it (D) 0 (E) (F) (E) f() (F) f() + (G) (H) undefined (G) f() (H) f() Eample 7: Use the graph of f() to find y = f() f() f() f() + (D) f() (A) (B) 0 (C) No it (D) (E) (F) (E) f() (F) f() + (G) (H) (G) f() (H) f() Eample 8: The figure is the graph of f() on the interval [,]. Find all true statement(s). f() = 0 (B), (D) and (F) f() = f() does not eists + f() = 0 (E) f() = 0 (F) a f() eists for all < a <

Eample 9: Use a table of values for the function to estimate the it if it eists: 9 + Let f() = 9 +. Step : Find the values of f() near =..9.99.999.00.0. f().0768 0.007 0.0007 0.0007 0.007 0.07 f() approaches 0 f() approaches 0 Step : Find the value to which f() is close as is close to, if any: By Step, f() = 0 and f() = 0 : + The left it is equal to its right it. Therefore, 9 + = 0 Eample 0: Use a table of values for the function to estimate the it if it eists: Let f() =. Step : Find the values of f() near =..9.99.999.00.0. f() f() approaches f() approaches Step : Find the value to which f() is close as is close to, if any: By Step, f() = and f() = : + The left it is NOT equal to its right it. Therefore, does not eist Eample : Use a table of values for the function to find the it if it eists: 0 sin ( π ) Step : Find the values of f() = sin ( π ) near = 0. 0 7 9 9 7 f() f() oscillates f() oscillates

Step : Find the value to which f() is close as is close to 0, if any: By Step, f() oscillates between and from both sides. Then f() and f() do not eist. 0 + 0 sin 0 (π ) does not eist π f( ) = sin( ) Eample : Use a table of values for the function to find the it if it eists:.9.99.999.00.0. y =.9.99.999.00.0. Eample : Use a table of values for the function to find the it if it eists: 9 + 6..0.00.999.99.9 y = 9 + 6. 6.0 6.00.999.99.9 Eample : Use a table of values for the function to find the it if it eists: 0 cos ( π ) DNE 0 y = cos ( π ) Eample : Use a table of values for the function to find the it if it eists: ( ) 0.9 0.99 0.999.00.0. ( ) 00 0000 000000 000000 0000 00

DEFINITION: (Infinite it definition) Assume that y = f() is defined in some open interval containing a. ) If f() is increasing without bound for all sufficiently close to a and < a without letting = a, then we said that the left it of f() is as is close to a and write this as a f() = ) If f() is decreasing without bound for all sufficiently close to a and < a without letting = a, then we said that the left it of f() is as is close to a and write this as f() = a ) If f() increasing without bound for all sufficiently close to a and > a without letting = a, then we said that the right it of f() is as is close to a and write this as a + f() = ) If f() is decreasing without bound for all sufficiently close to a and > a without letting = a, then we said that the right it of f() is as is close to a and write this as f() = a + ) If f() is increasing without bound for all sufficiently close to a without letting = a, then we said that the it of f() is as is close to a and write this as a f() = 6) If f() is decreasing without bound for all sufficiently close to a without letting = a, then we said that the it of f() is as is close to a and write this as a f() = Infinite Left Limit y = f() a y = f() a f() = a f() = a Infinite Right Limit y = f() a y = f() a f() = a + f() = a + Infinite Limit y = f() a y = f() a f() = a f() = a

DEFINITION: The line = a is called a vertical asymptote (V.A) of the function y = f() if it satisfies one of those; f() = a f() = a f() = a + f() = a + f() = a f() = a Eample 6: Find the it by using the graph of y = f() in the figure if it eists f() f() (E) f() + f() f() (F) f() + (G) f() (I) f() + (H) f() (J) f() y = f() (A) (B) (C) DNE (D) (E) (F) (G) (H) (I) (J) Eample 7: Find the it by using the graph if it eists: + (A) f() is close to as is close to from the left: = Therefore, y = has a V.A =. (B) f() is close to as is close to from the right: = Therefore, y = has a V.A =. (C) The left it and the right it of f() at = are different. So, does not eist (Mathematicians frequently abbreviate does not eist as DNE)

Eample 8: Find the it by using the graph if it eists (use your calculator!): 0 0 (E) (G) tan() π 0 + 0 + ln() (F) (H) tan() + π (A) (B) (C) (D) (E) (F) DNE (G) (H) Eample 9: Sketch the graph of a function f that satisfies the following conditions: () f() =, + () f() =, () f() = There are infinite many solutions. We are going to draw a solution; FIRST CONDITION SECOND CONDITION THIRD CONDITION Put the boundary points as open points since the it is the approaching value of f near =, not f() Eample 0: Sketch the graph of a function f that satisfies the following conditions: () + f() = () f( ) = 0 y = f() () f() = () f() = () f() = (6) f() = Eample : Sketch the graph of a function f that satisfies the following conditions: () f() =, + () f() =, y = f() () f() =

6 PRACTICE PROBLEMS. Use the graph of y = f() to find f() if f() = +, f() if f() = {, = (A) (B) (C) DNE (D) y = f() y = f(), f() if f() = { +, > f() if f() = + y = f() y = f(). Use the graph of f() to find (A) f() f() + (B) (C) y = f() f() (D) f() (D) Undefined (E) (E) f() (F) f() + (F) (G) (G) f() (H) f() (H) 0. Use the graph of f() to find (A) f() f() + (B) (C) f() (D) f() (D) (E) (E) f() (F) f() + (F) (G) DNE y = f() (G) f() (H) f() (H). Use the graph of f() to find (A) y = f() f() f() + (B) (C) f() (D) f() (D) Undefined (E) (E) f() (F) f() + (F) (G) DNE (G) f() (H) f() (H)

7. Use the graph of the following function y = f() in the figure to find f() f() (E) 0 f() (G) 0 f() (I) f() (K) f() (M) f() f() + (D) f( ) (F) f() 0 + (H) f(0) (J) f() + (L) f() (N) f() (A) (B) (C) (D) (E) 0 (F) 0 (G) 0 (H) Undefined (I) (J) (K) DNE (L) (M) (N) 6. Use the graph of the following function y = f() in the figure to find f() f() f() (F) f() (G) f() (I) f() f() + (E) f() + (H) f() + (A) (B) (C) (D) (E) (F) DNE (G) (H) (I) 7. Use the graph of the following function y = f() in the figure to find 8 6 6 f() f() f() (F) f() (G) f() (I) f() f() + (E) f() + (H) f() + (A) (B) (C) DNE (D) (E) (F) (G) (H) (I) DNE 8 8. Find the it by using the graphic utility if it eists 0 0 + 0 (E) + (F) (G) 0 (H) 0 + (A) (B) (C) (D) (E) (F) DNE (G) (H)

8 9. Create a table of values for the function and use the result to estimate the it. 9 sin () 0 0 cos ( ) (A) 6 (B) (C) (D) DNE 0. Sketch the graph of a function f that satisfies all the given conditions: ) f() =, ) + f() = ) f(0) = ) 0 f() = f(0). Sketch the graph of a function f that satisfies all the given conditions. + f() =, f() =, f() =, f() =, f( ) = 0 0 f() =, + f() =, f() =, f( ) = f() =, + f() =, f( ) = 0, f( ) =, f() =. Use the graph of f to find all values of c such that f() eists. 6 (, ) (, ) c y = f(). Use the graph of y = f() to find the its if it eists: f(f()) and f(f()). + f(f()) = f(f()) = 0 + y = f()

9 LECTURE - BASIC METHOD TO FIND LIMITS We are going to learn the actual definition of its in Lecture -7 and how to find the it by using the definition. Right now, we use the properties of it (it is induced by the definition) to find the it. PROPERTIES OF LIMITS: Assume that f() and g() are defined for any in some open interval containing a (without letting = a). Suppose that c is a constant and a f() = L, a g() = M eist. ) a (f() + g()) = L + M ) a (f() g()) = L M ) a (f()g()) = LM ) a ( f() g() ) = L M if M 0 ) a cf() = cl 7) a c = c n n 6) f() = L ( L 0 if n is even) a 8) a n = a n for any real number n 9) a p() = p(a) for a polynomial p() p() 0) a = p(a) q() q(a), q(a) 0 for polynomials p(), q() ) If h() is defined for any in some open interval containing L and L h() = T, then a h(f()) = T Eample : Let f() =, g() = and h() = [f() g()] [f() g()] = f() g() By ) and ) = () ( ) = 9 [g()] h( )+ [g()] h( )+ = [ g()] h( )+ = ( ) + = 9 By ), ) and ) Eample : Let f() = and g() =. [f() + g()] [ f() g()+ ] [ f()g() [g( + ) ] ] (A) (B) (C) (D) 6

0 Eample : (Substituting = a) Find the it if it eists ( + 7) ( + 7) = () () + 7 By 9) = 9 + + = ( ) ( ) First, check g( ) = ( ) 0 if g() = By 0) = = + + = + () + = 0 First, check g() = + 0 if g() = + By 0) = () By 9) = = = ( ) by ) where f() =, g() = Eample : Find the it if it eists ( + ) (E) + ( + (G) cos ( π ) ( ) ln e ) (F) ++ ( ) + (H) sec (π) (A) (B) (C) (D) (E) (F) 0 (G) (H) (I) 0 (J) 6 (I) + (J) < f(), f() = {, +,

THEOREM -: Let f() = g() on the interval (a, b) ecept c where a < c < b. If c g() eists, then Even if g() is undefined (it has a hole at = ), f() = g() c c Since the graph of f is close to f() = as, g() = f() = y = g() y = f() f() How to find the? c g() ) First, find f(c) g(c). f() If it is defined, = f(c) c g() g(c) ) When f(c) g(c) is of the form 0 0 Factor and simplify by Theorem -. Then substitute = c to find the it. If it has radicals, use conjugate to remove the factor c. Then substitute = c to find the it. ) When f(c) is other undefined form, g(c) Rewrite it as one fraction and simplify by Theorem -. Then substitute = c to find the it. For eample, consider g() if g() = +. For any, g() = + = ( )( + ) ( + ) y = By Theorem -, = + = ( ) = ( ) 6 =

Eample : Find the it if it eists = ( + )( ) ( ) f() = = 0 if f() = 0 Factor (+) 9 8 6 = ( + ) By Theorem - = () + By substituting = = Simplify ( + ) 9 = + 8 = ( + )( ) ( ) f() = (+) 9 = 0 if f() = (+) 9 0 Simplify the numerator Factor 8 6 = ( + ) By Theorem - = + By substituting = = 6 Simplify Eample 6: Find the it if it eists + 0 (E) / 6 (G) ( ) 0 (I) (K) +8 + (M) {ln( ) ln( )} + + ++ (F) + + 8 (H) ( ) (J) + (L) + 9 8 + (N) {log( 9) log( )} (A) (B) (C) 0 (D) (E) (F) (G) (H) (I) (J) (K) (L) (M) ln (N) log 6

Eample 7: Find the it if it eists: 0 9 0 9 = 0 ( ) = 0 ( ) ( )( + ) = 0 + = 0 + = + = sin () π sin() f(0) = 0 = = 0 if f() = 9 0 0 9 Factor By Theorem - By substituting = 0 Simplify sin () π sin() (sin() )(sin() + ) = π (sin() ) f ( π ) = sin (π ) = = 0 if f() = sin () sin(π ) 0 sin() Factor = π (sin() + ) By Theorem - = sin ( π ) + By substituting = π = + = Simplify Eample 7: Find the it if it eists: 0 0 (E) cos cos + 0 cos (G) + 0 sin cos π/ sin cos sin + sin π/ sin (F) cos cos 0 cos cos() (H) 0 cos() (A) (B) (C) (D) (E) (F) (G) (H)

Eample 9: ( 0 0 and case) Find the it if it eists: +9 0 0 + 9 = 0 ( + 9 )( + 9 + ) ( + 9 + ) ( + 9) 9 = 0 ( + 9 ) = 0 ( + 9 + ) = 0 + 9 + = 9 + f(0) = 9 0 + = 0 0 if f() = +9 by multiplying the conjugate of the numerator to both sides Simplify the numerator. By Theorem - By substituting = 0 = 6 + = ( )( + )( + + ) ( + )( + + ) = ( )( + )( + + ) ( ) f() = + = 0 0 if f() = + by multiplying the conjugate of the numerator to both sides Simplify the denominator. = ( + )( + + ) By Theorem - = ( + )( + + ) By substituting = = 8 Eample 0: Find the it if it eists: + +8 (E) + (G) 9 9 0 0 6 (F) + + (H) + (A) (B) (C) 6 (D) (E) 6 (F) (G) (H) 60

Eample : (complicate form) Find the it if it eists: + { } When it is a complicate fraction form, first simplify. + ( ) ( + ) = ( + ) ( + ) ( ) ( + ) = ( + ) ( + ) ( + ) = ( ) ( + ) ( + ) = ( ) ( + ) ( ) ( + ) = { } = 9 f() = = 0 + 0 if f() = + ( 0 + ) Find the least common multiple of denominators:( + ) By the distributive law Simplify By Theorem - By substituting = { = f() } = { = { = = { ( + ) ( )( + ) ( )( + ) } ( )( + ) } 0 0 + } By Theorem - is undefined if f() = Rewrite it as a fraction and simplify. By substituting = 0 ( + ) f(0) = 0 is undefined if f() = ( ) 0 + = 0 ( ( + ) ( + ) ( + ) ) = 0 ( + ) = 0 ( + ) Rewrite it as a fraction and simplify. By Theorem - = By substituting = 0

6 Eample : Find the it if it eits: (+h) 7 ( 7) h 0 h h 0 ( + h) 7 ( 7) h = h 0 + h 7 + 7 h h = h 0 h (+h) (+h)+ ( +) h 0 h Simplify the numerator. Reduce the fraction = h 0 = h 0 ( + h) ( + h) + ( + ) h = h 0 + h + h h + + h = h 0 h + h h h = h 0 h( + h ) h = h 0 ( + h ) Simplify the numerator. Factor the numerator Simplify = Limit at h approaches to 0 Eample : Find the it if it eists: + { 6 } 9 (E) ( ) + (G) ( ) + { } 0 { } (+) (F) ( ) (H) ( + ) 0 + (A) (B) (C) 6 (D) (E) (F) 9 (G) (H) 9 Eample : Find the it if it eists: (+h) ( ) h 0 h (+h) 7(+h) ( 7) h (A) (B) 6 7

7 PRACTICE PROBLEMS. Let f() = and g() =. Find the it if any (f() + g()) (f() g()). Let f() = and g() =. Find the it if any f() (f() + g()) (E) f(g()) (f() + ) ( f()+ +g() ) g() [f() g()] (F) g(f()) (A) (B) (C) (D) DNE (A) (B) (C) (D) 6 (E) 0 (F) 6. Using the graphs of the functions, find [f() + g()] 6 [f()g()] 8 (A) (B) g f 6 g f. Find each it if it eits, and justify each step. ( + ) 8+ + (E) {( )( + + )} + ++ + + (F) (G) csc ( π ) (H) π/ sin() (I) {sin() cos()} π/ (K) (M) cos π/ tan (O) π/ log 9 tan (Q) 0 [log log ( )] (J) {sec () tan ()} (L) sec ( π ) (N) log () (P) + 0 + 0 (R) (A) (B) (C) (D) 0 (E) 8 (F) (G) (H) (I) 0 (J) (K) (L) (M) (N) (O) (P) (Q) (R). Find each it if it eits, and justify each step. + 6 (A) (B) (C)

8 + (E) 6+8 + 8 (G) + 6 + (I) + (K) (+) 0 (M) (+) 9 0 (O) 7 +7 + 0 (+h) (Q) h 0 h (S) {(+h) (+h)+} ( +) h 0 h + (F) ++ (H) ( )( ++) (J) + (L) (N) +7 + (P) 6 {(+h) } ( ) (R) h 0 h (+h) (T) h 0 h (D) (E) (F) (G) (H) 7 (I) (J) (K) (L) (M) 6 (N) 7 (O) (P) (Q) 8 (R) (S) 6 (T) 6. Find each it if it eits, and justify each step. + 0 (E) +6 (G) + +8 8 (I) (K) + 9 9 + (F) (H) + (J) + 9 9 (L) (A) (B) 6 (C) (D) (E) 6 (F) 6 (G) (H) (I) (J) (K) (L) 7. Find each it if it eits, and justify each step. tan tan π/ tan + 8 (E) e e + 0 e (G) (I) cos π/ sin sin cos π/ tan cos cos + 0 cos 0 tan (F) 0 sin (H) (J) cos π/ sin() cos() π/ sin() (A) (B) (C) 6 (D) (E) (F) (G) (H) (I) (J)

9 8. Find each it if it eits. ( ) 0 ( + ) (E) + + ( ) + ( ) + (+) (F) 0 (A) 6 (B) 6 (C) (D) 9 (E) (F) 9. Let f() = + and g() =. Find each it if it eists. f() g() g(f()) (A) (B) 9 (C) 9 0. Let f() = + and g() = +. Find each it if it eists. f() g() g(f()) (A) (B) 8 (C) 8 f() f(a) f(a+h) f(a). Find and for: a a h 0 h (A) f() = +, a = (B) f() =, a = (C) f() =, a = 0 (D) f() =, a = (E) f() =, a = (F) f() =, a = (A) (B) 6 (C) (D) 8 (E) (F). Let f() =. Find the following it; + f() f() + f() f() 6 (A) (B) (C) (D) No it. Find the constant a such that f() where f() = + a : greatest integer function. denotes the a =. Find the it: denotes the greatest integer function 0 (E) + + + 0 + (F) + (A) (B) (C) (D) 0 (E) 0 (F) 0. Find f() if f() f() = 7 = 7 (A) 9 (B)

0 f() 6. If a polynomial f() satisfies =, find f( ). 0 6 8 ( 7. If a polynomial f() satisfies ) =, find f(). 6 ( )f() 8. Find f() where f() = + 0 + + + (+ ) +. (+ ) ( 9. Find f() if a polynomial f() safisfies )f() = 6. ( ) ( )f() 0. Find if f() =. 8 f(). If a polynomial f() satisfies =, find + {f()} f(). f() f(). Find f() 0 +f() f() if =. 0. Find a polynomial f() = a + b + c + d such that f() f() = and =. Find a polynomial f() = a + b + c + d such that f() = 7 + 7 f() = + f() 0 ( ) = and = ( ) f(). Let A(,) and P(a, b) (a > ) be points on the graph of y =. Let HP be a segment which is perpendicular to -ais and AQ be a segment which is parallel to -ais. If P moves to A through the graph of y, find the it of AQ as P approaches to A. PQ A P Q H 6. Let A(,0) and P be points on the graph of y =. If P moves from A to the origin O(0,0) through the graph of y, find the it of the measure angle APO as P approaches to O: P O (measure of APO). P y = O A

LECTURE - LIMITS AT INFINITE DEFINITION: The line y = L is called a horizontal asymptote of a function y = f() if either f() = L or f() = L horizontal asymptpte LONG TERM BEHAVIOR: The long-term behavior describes how a graph appears as the independent variable approaches infinity to the right ( increases) or to the left ( decreases). It depends whether the degree of the polynomial is odd or even and the sign of the coefficient of the highest order term, a n. The long-term behavior for all possible cases is: Let f() and g() be polynomials. f() g() = leading term of f() leading term of g() f() g() = leading term of f() leading term of g() THEOREM -: If n > 0 is a rational number, then = 0 if n n as y = = 0 if n n ± as Eample : Find each it and its horizontal asymptote (H.A) if it eists. ( ) ( ) = ( ) = Then, y = does not have any H.A. ( + ) By the long-term behavior ( + ) = ( ) = Then, y = + does not have any H.A. By the long-term behavior

Eample : Find the it and its horizontal asymptote (H.A) if it eists. + + = = 0 0 = + + + + Divide numerator and denominator by By theorem - and ± /n = 0, n > 0 ( )(+) ( )(+) Then, it has a horizontal asymptote y = + = + = = Divide numerator and denominator by By theorem - and ± /n = 0, n > 0 Then, it does not have horizontal asymptotes. + + + = + Divide numerator and denominator by = 0 By theorem - and ± /n = 0, n > 0 Then, it has a horizontal asymptote y = 0 ( )( + ) ( )( + ) + 6 = + + 6 = + = By distributive law Divide numerator and denominator by By theorem - and ± /n = 0, n > 0 Then, it has a horizontal asymptote y =

Eample : (Use the long term behavior) Find the it and its horizontal asymptote (H.A) if it eists. ( )( ) + ( )( ) + = + + = = = +7 +8 By distributive law By the long-term behavior Reduce and find the it 6 8 6 7 6 Then, it has a horizontal asymptote y = + 7 + 8 = By the long-term behavior = = Reduce and find the it Then, it does not have horizontal asymptotes. = = 6 8 6 7 6 6 = 0 Then, it has a horizontal asymptote y = 0 By the long-term behavior Reduce and find the it Eample : Find the it if it eists ( ) ( + ) (E) + 9 + (G) +6 + (I) ( )(+) + (K) {ln( + ) ln } + +8 (F) + (H) +7 ++ (J) {log(0 + ) log } (L) tan ( ) (A) (B) 0 (C) (D) (E) (F) (G) (H) 0 (I) (J) (K) ln (L) π

In the previous eample the infinity that we were using in the it didn t change the answer. This will not always be the case so don t make the assumption that this will always be the case. Let s take a look at an eample where we get different answers for each it. The square root in this problem won t change our work, but it will make the work a little messier. We use the following property for a natural number n; n = n = { n, < 0 n, 0 In this case, we divide numerator and denominator by t where t is the largest power of in the denominator. Eample : ( and case) Find the it if it eists: ++ 6 6 7 7 + + = + + = + + = + ++ 9 6 8+9 The largest degree in denominator is since = Divide each term by = since > 0 Simplify Since n = 0, n > 0 = = = = + 0 + + + + + The largest degree in denominator is since = Divide each term by = since < 0 Simplify Since n = 0, n > 0 =

6 6 7 7 6 = 6 6 = 7 = 7 7 6 7 7 7 The largest degree in denominator is Divide each term by Simplify Since n = 0, n > 0 = since > 0 = 7 9 6 8 + 9 96 = 6 8 6 + 9 9 8 = 9 = 9 The largest degree in denominator is Divide each term by 6 = since < 0 Simplify Since n = 0, n > 0 = 7 Eample 6: Find the it if it eists + 9 + + 8 (E) + + (G) + + (I) + ++ (K) +7+ + + 9 + 6 +8 +7 (F) 7+ + (H) ++ (J) 6 9 ++ (L) 6+ 9 + 8 (A) (B) (C) (D) (E) (F) 7 (G) (H) (I) (J) (K) (L) 6

6 Eample 7: ( case) Find the it if it eists: + + : case which MAY NOT be zero. We rationalize the numerator to find it. + = ( + )( + + ) + + = ( + ) + + = + + = + + = + = multiply numerator and denominator by the conjugate radical Simplify Divide each term by = since > 0 Since n = 0, n > 0 Eample 8: ( case) Find the it if it eists: + + : case which MAY NOT be zero. We rationalize the numerator to find it. + = ( + )( + + ) + + = ( + ) + + = + + = + + = 0 + = 0 multiply numerator and denominator by the conjugate radical Simplify Divide each term by = since > 0 Since n = 0, n > 0 Eample 9: Find the it if it eists ( + ) ( + ) ( + ) ( + 0 ) (A) 0 (B) (C) 0 (D)

7 Eample 0: ( 0 case) Find the it if it eists: ( + ) ( ) There are three cases if the approaching value is 0 : for eamples, First, we simplify the epression. ( + ) ( ) = ( ( + )( ) ( + ) ( + )( ) ) = ( ( + )( ) ) = ( 6 ) = ( 6 = ± a = a a { = 0 as ±. f() is close to 0 which is undefined as Rewrite it as one fraction Simplify ) Divide each term by = 0 Since n = 0, n > 0 ( ) = ( ( ) ( )( ) ( ) ( )( ) ) = ( + ( )( ) ) f() is close to 0 which is undefined as Rewrite it as one fraction Simplify = = + + + + Divide each term by = Since n = 0, n > 0 Eample : Find the it if it eists + ( ) + ( + ) ( + ) 0 + ( ) + (A) (B) 9 (C) 0 (D) 0

8 NOTE: Consider r and r. When r > When 0 < r < y = r, r > y = r, 0< r < r = 0 ; r = r = ; r = 0 Eample : Find the it if it eists. 7 8 (E) First, find the eponential form a whose base is biggest one in the denominator and divide a in each term. = 0 = ( ) = 0 = ( ) = 7 Since > and Since < and Since > and Divide each term by 8 8 7 ( = 8 ) ( 8 ) = 0 7 0 = 0 Since <, 8 < and 8 + + (E) = = 9 = ( = 9 ) Divide each term by Since < and

9 Eample : Find the it if it eists. First, find the eponential form a whose base is biggest one in the denominator and divide a in each term. Let A = = ( A A ) = ( A A ) +7 A as = + 7 = A A + 7 A A = A A 7 A A + A A A A ( = ) + ( A 7 ) A Since > and A Let A = A as Since <, < and 7 = 0 + 0 = 0 Eample : Find the it if it eists. ( ) (E) + (G) ( ) (F) 8 + 7 (H) + (A) 0 (B) (C) 0 (D) (E) (F) 0 (G) 0 (H) (I) (J) 0 e e (I) e + 7 + (J)

60 PRACTICE PROBLEMS Find the it if any ( ) + (E) + (G) (I) +7 ++8 ( 8) ( ) (F) ++ (H) + +6 (J) + + (A) (B) (C) 0 (D) (E) (F) (G) 0 (H) (I) (J) (K) 0 (L) (K) ++ (L) ++ (M) (+)( +) ( )( ) (N) 6 7 + (M) 7 (N) (O) 0 (P) 8 ( ) + (O) (Q) tan ( ) (S) {log (6 + ) log ()} (U) ( + ) (W) e + (Y) ( ) Find the it if any + ++ (E) ++ (G) + (I) + ++ (K) ++ 6 ( +) (P) ( 7) (R) {log ( 7+ )} (T) {log( + ) log(0)} (V) 8 + (X) e6 (Z) ( 7 ) + ++ (F) + (H) 9 + + (J) (L) + (Q) π (R) (S) (T) (U) 7 (V) (W) (X) 0 (Y) (Z) 0 (A) (B) (C) (D) (E) (F) (G) (H) 7 (I) (J) 0 (K) 0 (L)

6 Find the it if any ( + ) ( + ) ( + ) ( + + ) (A) (B) 0 (C) (D) Find the it if any. +7 (E) 7 +7 + 7 7 (F) (A) 0 (B) (C) (D) 0 (E) 0 (F) (G) (H) (G) ( ) (H) + Let P(A, B) be the intersection point of y = a (a > ) and y =. Find ( a )n+a ( a )n+ + n 6 Find the constant a if a + + +7 = 8 a = 7 Let f() = n+ +a n n +, > 0. Find the constant a if f() eists. a = 8 Find ( + ) where denotes the greatest integer function (.7 =,. = ) 9 Find a quadratic polynomial f() = f() + a + b such that = a =, b = 6 f() 0 Find (it eists) where f() f()+ +f() = 6 Let AP = CP = and PB =. If the measure of angle BPC is in the figure, the find the it (AB CB ). A B P C

6 A square is inscribed between y = a + a and -ais as shown in the figure. Find S where S is the area of this square (a > 0). a y y = -a +a Let ABC be a right triangle as shown in the figure such that AB =, B = 90, and CAB = 60. Let C, C, C, be points on the ray BC such that CC = AC = a, πa C = AC = a, C = AC = a,. Then, find n. n n A 60 a a B C a a C a C Given the graph of y = f(), let f () = (f f)(), f () = (f f)(),, f n () = (f n f)(). Find (A) (B) 6 y = f() y = n f n () n f n ()

6 LECTURE - PIECEWISE FUNCTION LIMIT AND INFINITE LIMITS THE LIMIT OF A PIECEWISE FUNCTION: To find the it of a piecewise function, use the left it and the right it at the given point. Only when f() = f() = L eists, then a a + f() = L a, Eample : Find f() if f() = { 8, < <., First, we find the left it and the right it at =. Therefore, Left it near = f() = = f() = Right it near = f() = + +(8 ) =, Eample : Find f() if f() = { +, < <., First, we find the left it and the right it at =. Left it near = f() = ( + ) = 6 Right it near = f() = + +() = 8 Since the left it is not equal to the right it, f() does not eist. Eample : Find the it if it eists f() if f() =, < { +, f() if f() =, < { +, (E) f() if f() = {,, = f() if f() =, < { + +, f() if f() = {,, > 8, < (F) f() if f() = {, (A) (B) 0 (C)DNE (D) (E)DNE (F)

6 NOTE: (ABSOLUTE VALUE) A, A < 0 A = { A, A 0 Eample : Find the it if it eists:. Step : Find the left it and right it. = Left it near = ( ) = since < 0 + = = Right it near = + ( ) since > 0 Step : Sine its left it is not equal to its right it, does not eist Eample : Find the it if it eists: + Step : Find the left it and right it by removing absolute value notation.. Left it near = + = ( ) since < 0 ( ) = ( + ) = 0 Right it near = + + ( ) = + ( ) = ( ) + = 0 since > 0 Step : since its left it is equal to its right it, + = 0 Eample 6: Find the it if it eists (E) + (F) + (A) 0 (B) (C) (D) DNE (E) DNE (F) 0

6 DEFINITION: The line = a is called a vertical asymptote of the function y = f() if it satisfies one of those; f() = a f() = a + f() = a vertical asymptote vertical asymptote f() = a f() = a + f() = a THEOREM -: Let n > 0 be an integer number and let a be a real number. n = even n = odd if if if if ( a) n +0 since ( a)n > 0 as a = + ( a) n a ( a) n +0 since ( a)n > 0 as a + = + ( a) n a + ( a) n 0 since ( a)n < 0 as a = ( a) n a ( a) n +0 since ( a)n > 0 as a + = + ( a) n a + y = y = a ( a) n = even n ( a) n n = odd a Eample 7 (± constant( 0) 0 + + ++6 case): Find the it if it eists: (A) We can find easily the it by using graph. + + + + 6 = Algebraically, to find the it + + + + 6 = + + ( + )( + ) = + + ( +0 ) Factor near = ( ) + is close to + 0 (positive) 0 + y = + + 6 6 6 = +

66 Then + + ++6 = : The vertical asymptote is = (B) To find the left it: = is close to 0 8 6 The vertical asymptote of y = is the line =. y = 6 (C) First, find left and right it: + ( ) = + ( ) = + Then, = + ( ) ( ) is close to, + +0 is close to ( ) +0, y = ( ) 6 6 The vertical asymptote is the line =. (D) First, find left and right it: + = + = Then, does not eist. is close to, + +0 is close to, 0 8 6 6 y = The vertical asymptote is = even if does not eist! Eample 8: Find the it if it eists. + (E) 0 (G) ( ) 0 + 0 (F) (H) 9 (A) (B) (C) DNE (D) (E) (F) (G) (H) DNE (I) 0 (J) (I) e ( ) (J) e ( ) +

67 Eample 0: Sketch the graph of an eample of a function f that satisfies all the given conditions. ) f() =, + ) f() =, ) f() =, ) f() = There are infinite many solutions. We draw a graph which satisfies the conditions; Step : By ), f has a vertical asymptote, =. By ), f has a horizontal asymptote, y =. By ), f has a horizontal asymptote,y =. Draw the asymptotes; Step : Draw the conditions of y = f() f() = + f() = f() =, f() = Step : Draw a graph of y = f()

68 Eample : Sketch the graph of an eample of a function f that satisfies all given conditions. ) f() =, + ) f() =, ) f() =, ) f() = 6 6 Eample : Sketch the graph of an eample of a function f that satisfies all given conditions. ) f() = f() = + + ) f() = f() = ) f() = 0, and ) f() = 0 Eample : Sketch the graph of an eample of a function f that satisfies all given conditions. ) f() =, 0 + ) f() =, 0 ) f() =, ) f() = and ) f() =. Eample : Find the constant k such that f() eists where f() = { k +, + k, < We can find the left it and the right it at = f() = + +(k + ) = k +, f() = ( + k) = + k Since f() eists, f() = f(). + k + = + k k = k =

69 Eample : Find the constant k such that a f() eists where k +, (A) a = ; f() = { k, < k +, (C) a = ; f() = { k, < + k, (B) a = ; f() = { + k, < (D) a = ; f() = { + k, >, < (A) 8 (B), (C), (D) Eample 6: Find the constants a and b such that +a = b 6 +8 = a ++ (A) Since the denominator is close to 0 and the it eists as is close to, it is 0 form case; 0 Therefore, So, a = and b = + a = 0 when = + a = 0 a = 0 a =. + ( )( + ) b = = = ( + ) = (B) It is the it at infinite; = 6 + 8 6 a + + = + 8 6 a + + = a = 6 a Therefore, a = 8 = 6 a a = Eample 7: Find the constants a and b if 8 = a a b (E) a +7 9 6+ = +a b (G) = = + +a+ = = +a+b (F) +9 = a + 7 a (H) + = + (A) a = (B) a = (C) a =, b = 6 (D) a =, b = (E) a = (F) a = 6 (G) a =, b = (H) a =

70 PRACTICE PROBLEMS. Evaluate the following its. + + (E) 0 { } (G) 0 { }. Find the it if f() = { +,, = 0 f() f(). Find the it if it eists., < f(); f() = {, +, < f() ; f() = {,, < (E) f() ; f() = {, =, > (G) f() ; f() = + + 6+9 (F) 0 + { } (H) (ln 9 ln ) f() f(), < f(); f() = {, f() ; f() = {,, = +, < (F) f() ; f() = {, =, > (H) f() ; f() = (A) DNE (B) (C) DNE (D) 0 (E) (F) 0 (G) DNE (H) ln 6 (A) (B) (C) 0 (D) (A) DNE (B) (C) 9 (D) DNE (E) (F) DNE (G) DNE (H) 0. Find the it if any 0 ( ) + + (E) 0 (G) ( ) (I) (K) π + csc ( ) ( ) + + (F) 0 ( + ) ( ) (H) (J) 9 + (L) 0 cot (A) (B) (C) (D) (E) (F) (G) (H) (I) (J) DNE (K) (L). Find the it if any 0 e 0 + ln() 0 e 0 e (A) (B) (C) (D)

7 6. Sketch the graph of an eample of a function f that satisfies all the given conditions. f() =, f() =, f() =, f() = 0, 6 6 7. Sketch the graph of an eample of a function f that satisfies all the given conditions. f() =, 0 + f() =, 0 f() =, f() = 6 6 8. Sketch the graph of an eample of a function f that satisfies all the given conditions. f() =, + f() =, f() =, f() = + c, c 9. Suppose f() = {. Find the constant c if f() eists. c, < k, k 0. Suppose f() = {. Find the constant k if f() eists. + k, < = 8 = k +,. Suppose f() = {. Find the constant k if +, < f() eists. k =. Find the constants a and b if + +a+ = + a + + = a+b = = +a+b (E) +a + (G) 0 ( ) = a +b 9 + = b (F) = a +b (H) a + = 8 a cos (I) = b (J) 0 tan ( + + a) = b (A) a = (B) a = 6 (C) a =, b = 8 (D) a =, b = (E) a =, b = (F) a =, b = (G) a = b = ± (H) a = (I) a =, b = (J) a =, b = 0 (K) a =, b = (L) a =, b = 6 (K) ( + + (a )) = b (L) ( a + b + ) =

7 f() 8. If = 0, find f(). 8. Let f() = a + b + c be a polynomial such that f() 9 Find f() = and f() + =.. Find a polynomial f() such that f() ++ = and =. f() = + f() 6. Let P(t, t) be a point on the graph of y = and let l be a line passing through P such that l is perpendicular to segment OP (O is the origin) as shown in the figure. If m is the slope of l, find (OP m ) where OP is the length of the segment OP. t l t O t P y = 7. Find where AB is the length of arc AB and AB is the length of segment AB in the AB θ 0 AB sector OAB. B 8. Let C be a unit circle. Let C, C,, C n be inscribed same size circles in the circle C such that C t and C t+ are tangent each other eternally for each t n, C n and C are tangent each other eternally as shown in the figure. Let S n be the sum of the areas of those n circles C, C,, C n. Then find n ns n. O θ A π Cn C C C C C

7 LECTURE - SQUEEZE THEOREM AND TRIGONOMETRIC LIMITS SQUEEZE THEOREM (= Sandwich Theorem= Pinching Theorem) If a h() = a f() = L and h() g() f() when is near a (ecept possibly at a), then g() = L a Eample : If f() + for all 0, find f(). Since f() is bounded by two functions near =, we can apply the squeeze theorem. By the squeeze theorem, f() =. f() ( + ) f() Eample : Find the it: 0 sin ( ) We know that sin ( ) 0 sin ( ). 0 sin ( ) sin ( ) Since = 0 and by the squeeze theorem, sin ( ) = 0. 0 0 Then 0 sin ( ) = 0 since sin ( ) sin ( ) sin ( ). Eample : Find f() if for all, (A) + f() + + (B) f() + (A) (B) f() = f() = Eample : Find the it if it eists: 0 + cos ( ) sin (A) 0 (B) 0

7 TRIGONOMETRIC LIMITS (a 0, sin(a) 0 a = = sin(a) 0 a 0 0 case) tan(a) 0 a sin(a) cos(a) (Actually = if A 0 as 0) = 0 0 A 0 a = = tan(a) 0 a sin() Eample : Prove that = 0 Step : When > 0 and is radian Since sin > 0, < OAB < sector OAB < OAT r sin < r < r tan < sin cos sin < < tan > sin Since cos as 0, by sandwich theorem Step : When < 0, let = θ. Step : By step and, sin() = 0 + > cos sin() sin( θ) = = 0 θ 0 + θ θ 0 + sin() = 0 odd function sin(θ) θ O sin(θ) = = θ 0 + θ r r B A T r tan() cos Eample 6: Prove that = 0 0 cos 0 = 0 ( cos )( + cos ) ( + cos ) cos = 0 ( + cos ) sin = 0 ( + cos ) sin = ( 0 sin + cos ) sin = 0 sin 0 + cos = 0 = 0 Multiply conjugate of numerator to both numerator and denominator. sin + cos =

7 Eample 7: Find the it if it eists: sin() 0 sin() 0 sin() sin() sin() = 0 sin() = 0 = 0 = 0 sin() ( 0 0 form) sin() 0 ( 0 0 form) = = sin() 0 tan() 0 Until you find the it, do not forget to write the it symbol 0 sin() = 0 = 0 = 0 = sin() ( 0 0 form) = sin() 0 ( 0 form) 0 Until you find the it, do not forget to write the it symbol 0 tan() tan() = 0 = 0 = 0 = ( 0 0 form) tan() = 0 ( 0 0 form) Until you find the it, do not forget to write the it symbol Eample 8: Find the it if it eists. 7 0 sin() tan() 0 sin() (E) 0 6 tan(8) (G) 0 sin() (I) π sin() sin() 0 sin() cos 0 (F) 0 sin() tan() (H) 0 tan() tan(6) (J) 0 tan(7) (A) (B) (C) 7 (D) 0 (E) (F) (G) (H) (I) (J) 6 8 π 6 7

76 Eample 9: Find the it if it eists. sin( ) sin( ) sin( ) = ( + ) ( )( + ) sin( ) = ( + ) + tan( ) ( 0 0 form) cos() 0 = ( + ) sin( ) = since 0 as = + tan( ) = ( )( + ) tan( ) = = ( )( + ) tan( ) ( + ) tan( ) ( 0 form) 0 Factor Divide the numerator and the denominator by tan( ) = since 0 as = 0 cos() cos() = 0 = 0 ( 0 0 form) cos() 0 = 0 since 0 as 0 Eample 0: Find the it if it eists. sin( ) 0 cos 0 sin() sin() cos() cos 0 0 sin() cos() (E) 0 sin() sin(sin ) (G) 0 (F) 0 tan() cos() cos (H) 0 (A) 0 (B) 0 (C) (D) 0 (E) (F) (G) (H)

77 PRACTICE PROBLEMS. Find the it if any. sin() 0 0 sin(7) sin(0) (E) 0 sin() tan() (G) 0 tan() sin() cos() (I) 0 (K) cos 0 (M) 0 tan() sec() (O) 0 sin() sin(9) sin() sin() cos(π) (Q) / sin() (S) 0. Find the it (Sandwich Theorem) 0 cos ( ) sin tan() 0 0 sin(9) tan() tan(6) (F) 0 sin() sin( ) 0 sin( ) (H) (J) 0 sin cos() (L) 0 sin() sin() (N) 0 sin() sin(8) (P) 0 sin() (R) sin(π) (T) sin( ) 0 cos cos( ) (A) (B) (C) (D) 7 9 (E) (F) (G) (H) 6 9 (I) (J) DNE (K) 0 (L) 0 (M) (N) (O) 6 (P) 0 (Q) (R) π (S) 0 (T) 0 (A) 0 (B) 0 (C) 0 (D) 0. If 9 f() + 7 for all 0, find f(). 7 f(). Find if f() for all 0 0 + +. Find f() if 9+ sin f() sin for all 0 f( ) 6. Find 8 if a polynomial f() satisfies 0 f() =. 7. Find the its. π sin( ) 0 π cos( ) 0 cos() cos() cos +cos 0 0 (A) 0 (B) (C) 8 (D) (E) (F) 6

78 (E) 0 sin() sin (F) 0 cos() cos( ) sin( ) (G) (I) tan π/ sin cos (K) 0 cos() cos (H) 0 sin (J) sin 0 +tan sin sin (L) 0 cos() (G) (H) 0 (I) (J) (K) (L) 0 (M) (N) (M) 0 sin(sin()) sin( ) (N) 8. Let P n = ( n, y n ) be the intersection point of y = n and y = tan(), where nπ π < π < nπ + π for all natural number. Find n. n n 9. Let f(n) = 0 sin +sin()+sin()+ +sin(n) Find the value of a if f() + f() + + f() = a for all natural number n. a = 0. Find a and the natural number n if 0 tan sin n = a (a 0). a =, n =. Let P be a point on the graph of a unit circle + y = and R a point on -ais in the figure. Let Q be a point on the graph of y = e. Let QP be parallel to -ais and let the segment QR be parallel to y-ais. If the angle of ROP (O is the origin) is θ (0 < θ < π ) and the area of the triangle ROT is S (T is the intersection of PO and QR ), the find S(θ). θ 0 + θ y = e Q P T + y = O θ R. Let P be a point (in the first quadrant) on the graph of y =. Let A(,0),B(0,), and O(0,0) be points in the plane. If the measure of angle APO is θ, find the it PB θ π 6 BO θ π/6 y = P θ A O B

79. Let ABC be an equilateral triangle such that ABC = ACB = θ and the segment BC =. Let C be a circle with the center O which is inscribed in ABC and intercepts D and E S(θ) in the figure. If the area of ODE is S, find θ 0 + θ A D E B θ O C. Let a unit circle be inscribed in ABC and let ADC be a triangle such that BAC = θ and ACD = θ. Find θ 0 +{θs} where S is the area of a triangle BCD. C θ θ A θ B D. Let P be a point (in the first quadrant) on the graph of y =. Let A(,0),B(0,), and O(0,0) be points in the plane. Let OS, QR, PQ be perpendicular to AP, OP, AB, repectively. If the area of ASQ is f(θ) and the area of PQR is g(θ) where θ is the measure of angle PAB, find θ f(θ) θ 0 + g(θ) P 8 S R A θ O Q B 6. Let A n be the area of the regular n-gon inscribed in the unit circle, and let B n be the area of the regular n-gon whose inscribed circle has radius. (A) Show that A n < π < B n (B) Show that A n = n sin (π n ) (C) Show that B n = n tan ( π n ) (D) Evaluate n A n and n B n This is the picture of A, B 6

80 LECTURE -6 CONTINUITY DEFINITION: Let f be a function defined on an open interval I containing a. ) f is continuous at = a if a f() = f(a). ) f is left continuous at = a if f() = f(a). a ) f is right continuous at = a if f() = f(a). a + ) A function is continuous on an interval I if it is continuous at every point in the interval I. In other words, f is continuous at = a if it satisfies the following three conditions; a f() eists. (B) f(a) is defined. a f() = f(a) DEFINITION: Let f be a function defined on a closed interval [a, b]. Then f is continuous on [a, b] if (A) f is continuous at the open interval (a, b) f() = f(b). b f() = f(a). a + DISCONTINUITY CASES: Removable discontinuity: f() = f() a a + Jump discontinuity: a f(), f() are eists but not same + a y = f() y = f() y = f() Infinite discontinuity a f() = ± or f() = ± + a Other discontinuity: We do not deal with this right now y = f()

8 Eample : Determine whether the function f is continuous at =,, 0,,. If not, determine what kind of discontinuity at the given points. = : infinite =, : jump = : removable CONTINUITY RULES: If f and g are continuous at a and c is a constant, then the following functions are also continuous at a: ) f + g and f g, ) f g, ) cf ) f g if g(a) 0 COMPOSITION: If f is continuous at a and f is continuous at g(a) then its composition (f g)() = f(g()) is continuous at a. f(g()) = f ( g()) a a THEOREM -7 ) Any polynomial is continuous everywhere ) Any rational function is continuous on its domain ) Rational functions, Radical functions, Trigonometric functions, Eponential functions, and Logarithmic functions are continuous on its domain. Proof: A polynomial is of the form where a n, a n,, a, a 0 are constants. Since b a m = a m and b m = b m for any m, A rational function is of the form R() = g() h() f() = a n n + a n n + + a + a 0 f() = a nb n + a n b n + + a b + a 0 = f(b) b b where g() and h() are polynomials. g() g() R() = b h() = b h() = b g(b) h(b) = R(b)

8 Eample : Find the domain of the following functions and interval(s) on which the function is continuous. (A) f() = (B) f() = 6 Solution (A) Since f is a rational function, the domain of f is (, 0) (0, ). By theorem -7, it is continuous on its domain. So, (, 0) (0, ) is the intervals on which the function is continuous. (B) Since f is a radical function, the domain of f is 6 0 6 0 ( )( + ) 0 By Theorem -7, it is continuous on its domain. So, [,] is the interval on which the function is continuous. Eample : Find the domain of the function f and interval(s) on which it is continuous. (A) f() = + (B) f() = (C) f() = + (E) f() = sin (D) f() = ln( ) (F) f() = cos (A) (, ) (B) [, ) (C) (, ) (, ) (D) (, ) (E) (, ) (F) [,] Eample 6: Determine whether the function f is continuous at =. (A) f() = (B) f() = (C) f() = + (D) f() = {,, = (A) Continuous (B) Continuous (C) Not continuous (D) Not continuous Eample 7: Is f() = + continuous at =? First find the left it at =, f(), and the right it at =. Left it near = f() Right it near = f() = f() = + Since all three values are same, f() = + is continuous at =.

8 Eample 8: Eplain why the function f() = {,, = is discontinuous at =. First find the left it at =, f(), and the right it at =. Left it f() Right it 0 = since 0 = + since 0 0 This is not continuous at = : Since the left it is, it is infinite discontinuous at =. Eample 9: Eplain why the function h() = { +, < 7, = is discontinuous at =. Determine whether the, > discontinuity is a removable discontinuity, a jump discontinuity, or an infinite discontinuity. First find the left it at =, f(), and the right it at =. Left it f() Right it ( + ) = 7 +( ) = 6 So, it is not continuous at = : Since the left it is equal to the right it and f() is not equal to the right it, the function is removable discontinuous at =., < Eample 0: Eplain why the function f() = { is discontinuous at =. Determine whether the, discontinuity is a removable discontinuity, a jump discontinuity, or an infinite discontinuity. First find the left it at =, f(), and the right it at =. Left it f() : Right it ( ) = = + 6 So, it is not continuous at = : Since the left it is NOT equal to the right it, the function is jump discontinuous at =.

8 Eample : Decide whether f is continuous at =. If it is not, decide whether the discontinuity is a removable discontinuity, a jump discontinuity, or an infinite discontinuity. (A) f() = + (B) f() =, < (C) f() = {,, < (E) f() = { +, (G) f() = +, < (D) f() = {, =, >, < (F) f() = {, =, > (H) f() = + (A) Continuous (B) continuous (C) Jump (D) Removable (E) Continuous (F) Jump discontinuous (G) Infinite discontinuous (H) Continuous. a +, < Eample : Find the constant a if f() = { is continuous on the interval (, )., Since f is continuous on (, ), f is continuous at =. Left it f() Right it f() = ( a + ) = a + f() = + Since f is continuous at =, all three values (its left it, f(), and its right it) are same: a + = a = Eample : Find the constant a if f() is continuous on the interval (, ). a, < (A) f() = {, (C) f() = { +a+ +,, = a +, < (B) f() = { +, (D) f() = { +6+a +,, = (A) (B) (C) 9 (D) (E) (F) 6 (E) f() = { a, < a, a, < (F) f() = { a, Eample : Determine whether f() = { sin (), < 0 is continuous on the interval (, ). 0, 0 We only check that f() is continuous at = 0 since f() = sin ( ) is continuous on the interval (, 0) and f() = 0 is continuous on the interval (0, ). By Eample, it is continuous at = 0. So, f is continuous on the interval (, ).

8 PRACTICE PROBLEMS. Determine the intervals on which the following function are continuous. (A) f() = (C) f() = + (E) f() = + (G) f() = (I) f() = (K) f() = {,, = (M) f() = (B) f() = + (D) f() = ln( ) (F) f() = sin + + (H) f() = (J) f() = + sin (L) f() = tan() (N) f() = +. Eplain why the following graph of a function y = f() is not continuous at = (A) (B) (C) (A) (, ) (B) (, ) (,) (, ) (C) (, ) (D) (, ) (E) (, 0) (0, ) (F) (, ) (, ) (G) [, ) (H) (, ) (I) (, ] (J) (, ) (K) (, ) (, ) (L) Ecept = π + nπ for any integer n. (M) (, ) (N) [,0) (0, ) (A) Jump discontinuity (B) Removable discontinuity (C) Infinite discontinuity. Find the points at which f has discontinuous. For each point, state the discontinuity that are involved. (A) (B) (C) (A) =,0 (B) =, (C) =,, y = f() y = f(). Determine whether the function is continuous at the indicated point. If not, determine whether the discontinuity is a removable discontinuity, a jump discontinuity, or an infinite discontinuity. (A) Continuous (B) Jump discontinuous (C) Removable discontinuous

86 (A) f() = { +, < +, <,, = (B) f() = {, =, =, > (C) f() =, = (D) h() = {, 0, =, = +, <, < (E) f() = {, =, = (F) f() = {, > +,, = (G) f() =, = (H) f() =, =. Find the constants a and b given that f is continuous on the interval (, ). + a, < (A) f() = { a, (B) f() = { (C) f() = { a, < a +, +, a, = (D) f() = { a, < ( + )a, (D) Infinite discontinuous (E) Removable discontinuous (F) Jump discontinuous (G) Infinite discontinuous (H) Jump discontinuous (A) a = (B) a = 6 (C) a =, (D) a =, (E) a =, (F) a =, (E) f() = { a a, < a +, (G) f() = { (I) f() = { (K) f() = { +a+b +a+b + a +a, 7, =, < +,, a, = a (F) f() = { a, < a, (H) f() = { (J) f() = { +a+b +a+b, a, =, +, > sin, < 0 (L) f() = { a +, 0 (G) a =, b = (H) a = 9, b = (I) a = 8, b = (J) a =, b = (K) a = (L) a = 6. Let f() = { +a b,. Find constants a, b given that f is continuous on (, )., = a =, b = 7. Using the graph of y = f(), which of the following statement is(are) true? f() = y = f() f() = f(f()) eists 0 f(f()) = (E) f(f()) = (F) f() is discontinuous at two points D, F

87 8. Using the graphs of functions, determine whether following is continuous or not on the interval [,]. y = f() y = g() (A) Discontinuous at = 0 (B) Discontinuous at = 0 (C) Discontinuous at = 0 (D) Continuous (A) f() (B) f() g() (C) g(f()) (D) f(g() ) 9. Let f() = and g() = + 6. Which of the following is continuous on (, )? D (A) f() + g() (B) f() g() (C) f() g() (D) f(g()) (E) g(f()) (F) f(f()) 0. Find the constants a and b if f() = n+ +n +a+b is continuous on (, ) for n n + any real number. a =, b =. Find the -value(s) at which f() = n n ++ n + is discontinuous. = ±. Let f() be a continuous periodic function on the interval (, ) with period such that, < f() = {. Find the constants a and b. a + b, a = 6, b = 0. Let f() be continuous on the interval (, ). Find f() if ( )f() = + + a.. Find all values of at which f() = is discontinuous. = ±, ±, 0. Given the graph of f() on the interval (,), let g() = f() + f( ) on the interval (,). Find all true statements. 0 f() eists 0 g() eists (C) g() is continuous at = y = f() - - B, C

88 0, is rational 6. Find all values of (if any) at which f() = {, is irrational None 7. Given the graphs of f and g, decide whether the functions are continuous at = 0. (A) Discontinuous (B) Continuous (C) Discontinuous y = f() y = g() (A) f() + g() (B) f() g() (C) g(f()) 8. Let f() be a function defined on all real numbers. Determine whether the statement is true or false. f() = f() f() (B) If =, then f() = 0 0 0 f() (C) If = 8, then = f() (A) False (B) True (C) True

89 LECTURE -7 DEFINITION OF LIMIT DEFINITION: Let f be a function defined on an open interval containing a (ecept possibly at a) and let L be a real number. The statement means that for each ε > 0 there eists a δ > 0 such that if f() = L a 0 < a < δ, then f() L < ε Eample : Given the it ( ) =, find δ such that ( ) < 0.0 whenever 0 < < δ. To find the interval of, We are going to simplify ( ) < 0.0 ( ) < 0.0 6 < 0.0 < 0.0 < 0.0 = 0.00 So, if we choose any δ such that δ 0.00, then it satisfies ( ) = < δ 0.0. Eample : Given ( ) =. Find the greatest value of (the threshold) δ such that ( ) < 0.0 whenever 0 < < δ 0.00 Eample: Given ( + ) = 7. Find the greatest value of (the threshold) δ such that ( + ) + 7 < 0.0 whenever 0 < + < δ. 0.00 Eample : Given ( ) =. Find the greatest value of (the threshold) δ such that ( ) < 0.0 whenever 0 < + < δ. 0.00