Calculus I (part 1): Limits and Continuity (by Evan Dummit, 2016, v. 2.01)

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1 Calculus I (part ): Limits and Continuity (by Evan Dummit, 206, v. 2.0) Contents Limits and Continuity. Limits (Informally) Limits and the Limit Laws One-Sided Limits Continuity Limits at Innity, Innite Limits Limits and Continuity In this chapter, we introduce the fundamental idea of a it, which captures the behavior of a function near a point of interest. Our primary interest in its is to establish the denition of a continuous function, and to lay the technical groundwork for the denition of the derivative.. Limits (Informally) Informally, a it is something that captures the local behavior of a function. Saying that the function f() has the it L as approaches some value a means that as gets really close to a, f() gets as close as we want to L. This idea may seem obvious or stupid at rst (especially given that we have described it in rather vague and imprecise terms), but in fact it is more subtle than it might seem. Eample: Consider the function f() = as approaches 9. As approaches 9, f() should approach 3, based on our knowledge of the square root function. We can compute f() for some values of close to 9: f() And indeed, the values approach f() Eample: Consider the function g() = 2, dened for, as approaches. Here is a short table of values: g() f() As gets close to, it looks like g() is approaching the value 2. We can justify this with the following calculation: when we can write g() = since cancellation is allowed when the thing being cancelled is nonzero. So as gets close to, it makes sense that g() gets close to 2. Eample: Consider the function h() = sin(), dened for 0, as approaches 0. ( )( + ) ( ) = +,

2 Here is a short table of values for positive : h() Note that h() = h( ), so we will see the same behavior for negative as for positive. As gets close to 0, it looks like h() is approaching the value. However, unlike with g() in the eample above, there is no obvious algebraic manipulation we can perform that would eplain why h() is approaching as approaches 0. It certainly seems very plausible that as takes even smaller values, h() will continue getting closer and closer to 0, but at the moment we do not really have any way to justify this statement. ( ) Eample: Consider k() = cos, dened for 0, as approaches 0. Here is a short table of values for positive : k() Note that k() = k( ), so we will see the same behavior for negative as for positive. From these values it seems as though k() is approaching the value as approaches 0. But it we compute k() for even smaller, we see that this table was misleading: k() It appears that there is no pattern to the numbers as approaches 0, even as becomes very small. In hindsight, we should not have epected these numbers to have any kind of nice pattern: after all, as approaches 0, ( ) becomes very large (positive or negative) and so the function cos will oscillate wildly between the values and + near = 0. Here is a graph of the function y = k(), illustrating this unpredictable behavior near = 0: As we can see from these eamples, sometimes functions seem to have nice iting behavior near particular values (even if those values are not actually in the domain of the function), and other times they do not. Calculations like the ones we made in the eamples can only take us so far. For eample, for h() = sin(), how do we know that as we consider smaller and smaller values of that k() will continue getting ( closer ) to? After all, we made an erroneous guess for the very similarlooking function k() = cos : it looked initially like the function values were approaching as approached 0, but in fact they don't. 2

3 To reiterate: how can we ever know that we've gotten close enough to conclude that the value tends to what our intuition says it does? We need something else (namely, a rigorous denition) that will allow us to see that our intuition about iting values is correct..2 Limits and the Limit Laws The formal denition of the it allows us to back up our intuition with rigorous proof. Denition: A function f() has the it L as a, written as f() = L, if, for any ɛ > 0 (no matter how small) there eists a δ > 0 (depending on ɛ) with the property that for all 0 < a < δ, we have that f() L < ɛ. The symbols δ and ɛ are the lowercase Greek letters delta and epsilon (respectively). Their use in the denition of the it is traditional. Also recall that the notation means the absolute value of, and denotes the distance from to zero. Remark: This formal denition is very opaque. It takes practice and eperience to become comfortable with what the denition means, and to see why it really does match the intuition of how a it should behave. One way to think of this denition is as follows: suppose you claim that the function f() has a it L, as gets close to a. In order to prove to me that the function really does have that it, I challenge you by handing you some value ɛ > 0, and I want you to give me some open interval (a δ, a + δ) on the -ais containing a, with the property that f() is always within ɛ for in that interval, ecept possibly at a. If f() really does stay close to the it value L as gets close to a, then, no matter what value of ɛ I picked, you should always be able to answer my challenge with an interval around a, because the values of f() should stay near L when is near a. We will not use the formal ɛ δ denition to calculate its, as it is quite cumbersome even for very simple functions. Instead, we will state some properties that its obey that can be used to reduce complicated its to simple ones whose values we know. Each of these properties can be justied using the formal denition, but the proofs are complicated. Furthermore, the details of the it laws' proofs are not terribly important to understanding what the laws say and how they are used, which is why they are not given here. Theorem (Limit Laws): Let f() and g() be functions satisfying f() = L f and g() = L g. Then the following properties hold: Basic its: If a and c are any constants, then c = c, and = a. The addition rule: [f() + g()] = L f + L g. The subtraction rule: [f() g()] = L f L g. The multiplication rule: [f() g()] = L f L g. Note that the multiplication rule yields as a special case (when g() is identically equal to a constant c) the constant-multiplication rule: [c f()] = c L f, where c is any real number. [ ] f() The division rule: = L f, provided that L g is not zero. g() L g The eponentiation rule: [ f() b ] = (L f ) b, where b is any positive real number. (It also holds when b is negative or zero, provided L f is positive, in order for both sides to be real numbers.) The inequality rule: If f() g() for all, then L f L g. 3

4 The squeeze rule (also called the sandwich rule): If f() g() h() and f() = h() = L (meaning that both its eist and are equal to L) then g() eists and also equals L. Eample: Evaluate t t 3 3t + t 3 formally using the it laws. First, we use the subtraction rule to see that (t 3) = (t) (3) = 3 = 2, where we used t t t the two basic its at the end. Net, we use the multiplication rule to see that (3t) = 3 (t) = 3, and the eponentiation rule to t t see that (t 3 ) = (t) 3 =. t t Then we can use addition and subtraction to see that (t 3 3t + ) = (t 3 ) (3t) + () = t t t t 3 + =. t 3 3t + t (t3 3t + ) Finally, we can use the division rule to see that = t t 3 (t 3) = 2. t Note that this is the result we would have gotten if we had just plugged in t = to the function. Eample: Evaluate 0 2 cos ( ). We cannot use the multiplication rule here, because, although = 0, 0 and the multiplication rule requires both its to eist. 0 cos ( ) does not eist, We use the squeeze theorem instead: because cosine ( is ) always between and + and 2 is always nonnegative, we have the inequalities 2 2 cos 2. Then because ( 2 ) = 0 and 2 = 0, which both also follow easily from the it laws and the 0 0 ( ) basic its, the squeeze theorem dictates that 2 cos = 0. 0 In general, when evaluating its using the it laws, a reasonable procedure is rst to try plugging in the iting value to the function. If plugging in yields a numerical value, then the it can usually be evaluated by direct applications of the it laws. (Using the it laws in this way is essentially bookkeeping.) If an indeterminate epression like 0 is obtained, it is often necessary to perform some kind of algebraic 0 manipulation rst. t + 2 Eample: Evaluate t 2 t 2 4 formally using the it laws. If we try plugging in, we see that the numerator and denominator are both zero, suggesting that we should try to simplify the epression rst. Here, we can factor the denominator: t 2 4 = (t 2)(t + 2). t + 2 t 2 4 = t + 2 t 2 (t 2)(t + 2) = t 2, where we cancelled the common fac- t 2 Then we can write t 2 tor. Finally, the subtraction and division laws give us t 2 t 2 = 4. Remark: The use of the it allows us to say something about the function f(t) = t + 2 near t 2 t = 2, 4 even though the function (as written) is not dened there. 4

5 3 Eample: Evaluate 9 9. If we try plugging in, we see that the numerator and denominator are both zero, suggesting that we should try to simplify the epression rst. Here, the denominator is a dierence involving square roots. A useful technique in this situation is to multiply by the conjugate: = 9 9 ( ) = ( 9)( + 3) = 9 9 ( 9)( + 3). Now we can cancel the common factor 9 and are left with the simple it = One-Sided Limits Sometimes, we are only interested in the behavior of a it as the parameter approaches from one side for eample, if we have a physical system and we only have data over some time interval and want to say things about how the system behaves at the ends of the time interval. This motivates us to dene one-sided its. The denitions are the following (although, as before, we will not use the denition to evaluate its): We say that a function f() has the it L as a from below, written as f() = L, if the following statement is true: for any ɛ > 0 (no matter how small) there eists a δ > 0 (depending on ɛ) with the property that for all δ < a < 0, we have that f() L < ɛ. We say that a function f() has the it L as a from above, written as f() = L, if the + following statement is true: for any ɛ > 0 (no matter how small) there eists a δ > 0 (depending on ɛ) with the property that for all 0 < a < δ, we have that f() L < ɛ. The only change in the denition is the restriction on the -ais: a two-sided it has 0 < a < δ, while the it from below has 0 < a < δ and the it from above has 0 < a < δ. Remark: Normal its are sometimes called two-sided its, to distinguish them from the two possible one-sided its. The term it, with no qualier, refers to a two-sided it. All of the it laws also hold for one-sided its, with all of the its being changed to the appropriate one-sided it. Eample: Find. 0+ We know that = 0, and so by the eponentiation rule we see = = Depending on one's preference regarding comple numbers, one might say that the two-sided 0 does not eist (because is not a real number for negative ). Here, the one-sided it avoids these diculties. Proposition: A two-sided it eists if and only if both one-sided its eist and have the same value. When the two one-sided its are equal, the two-sided it shares the same value. This result is more or less a restatement of the denitions of the one-sided its. In particular: a two-sided it won't eist if either of the one-sided its doesn't eist, or if the two one-sided its have dierent values. Some functions have dierent behaviors to the left and to the right of a given point: this will be detected by the one-sided its. Eample: Find the one-sided its 0+ and 0 5. Does the it 0 eist?

6 { We can see that = if > 0, so the graph of this function jumps from to + at = 0: if < 0 By our results on its of constants, we see that 0+ =, while 0 =. The (two-sided) it of this function at 0 therefore does not eist, since the one-sided its have dierent values. 2 if > 2 Eample: Let f() = 4 if = 2. Find f(), f(), and f() if < 2 Here is a graph of this function: When > 2, f() = 2, so Similarly, when < 2, f() = +, so 2+ (2 ) = 3 using the it laws. ( + ) = 3 using the it laws. 2 Since the two one-sided its are equal, the two-sided it eists and has the same value: 2 f() = 3. Notice that although f(2) = 4, the it 2 f() is equal to 3. This odd behavior is reected in the graph of y = f(), and eplains why there is a hole: the curves representing y = 2 for > 2 and y = + for < 2 both approach the point (2, 3) as 2. However, because f(2) is actually equal to 4, the point (2, 3) is missing from the graph of y = f(), as it has been replaced with the point (2, 4)..4 Continuity We will now discuss a very important property of functions which captures a notion of niceness. Denition: A function f() is continuous at = a if f() = f(a). A function f() is called everywhere continuous (or often, just continuous) if it is continuous at = a for all real numbers a in its domain. 6

7 Continuous functions are (generally speaking) much nicer than arbitrary functions. Continuous functions do not jump or have missing points, nor do they blow up to. A common interpretation of a continuous function is: its graph can be drawn without taking the pencil o of the paper. Most of the functions we will encounter (polynomials, trigonometric functions, eponentials and logarithms) will be continuous everywhere they are dened, ecept possibly at a small number of points. Proposition: Every polynomial is everywhere continuous. Proof: We want to show that p() = p(a), where p() is any polynomial in, and a is any real number. By the eponentiation rule, we know that n = already know that = a. ( ) n = a n, for any positive integer n, since we [ ] [ c n n] = c n a n, for any constant c n, By the multiplication rule, we know that (c n n ) = since we know that c n = c n. Finally, because every polynomial is just a sum of terms of the form c n n for some coecients c n, we can use the addition rule (repeatedly) to conclude that p() = p(a), where p() is an arbitrary polynomial. Proposition: The functions e, sin(), and cos() are continuous for all. The proof of this result requires using the technical denitions of these functions, either as its or as innite sums. Theorem: If f() and g() are continuous functions, then f() + g(), f() g(), f() g(), and f(g()) are continuous functions. If g() 0 then f() g() is also continuous, and if g() > 0 then [g()]a is continuous, for any real number a. If f() is one-to-one then f () is also continuous. The proofs for the sum, dierence, product, quotient, and eponentiation follow from the respective it laws. The proof for the composition uses the denition of continuity twice. The proof for the inverse function uses the intermediate value theorem (see below). Note that a quotient of functions is never continuous anywhere that the denominator is zero (since, although the it may eist, the quotient itself is not dened and therefore cannot equal the it). Proposition: Any quotient of polynomials p() is continuous everywhere it is dened, as are the trigonometric q() functions tan(), sec(), csc(), cot(), and the logarithm ln(). These results follow from the properties of continuous functions and the results that polynomials, sin(), cos(), and e are continuous. Eample: The function f() = is continuous at all 0, and is discontinuous at = 0. Theorem (Continuity and Limits): If g() is a continuous function, and f() = L, then g(f()) = g(l). In other words, we can move continuous functions through its without having to worry about changing the value. Technically, this description is slightly stronger than continuity: it is describing a continuous function of bounded variation. The problem is that the graph could have { an innite length even on a nite interval (and such a graph cannot be physically drawn using a sin(/) for 0 real pencil). One eample is f() =, which is continuous at = 0 by the squeeze theorem, but the length of 0 for = 0 the graph of y = f() on any open interval containing = 0 is innite. 7

8 The proof is very similar to the proofs of the it laws. Using properties of continuous functions can allow us to evaluate many complicated its with ease: ( Eample: Find tan e sin() ). π 2 cos() First, because e sin() and 2 cos() are both continuous functions, and the latter function is never e sin() zero (since cos() is between and ), the quotient is also continuous. 2 cos() Thus, π e sin() 2 cos() = e sin(π) 2 cos(π) = e 0 2 ( ) = 3. Then, since arctangent is continuous, we can move it through the it to write ( ) ( ) e sin() π tan = tan e sin() ( ) = tan 3 = π 2 cos() π 2 cos() 6. A fundamental property is that a continuous function has no jumps. The formal statement is as follows: Intermediate Value Theorem: If the function f() is continuous on the interval [a, b] then for any y between f(a) and f(b), there eists a value c in (a, b) such that y = f(c). Equivalently, f attains all values between f(a) and f(b) as goes from a to b. This theorem is essentially equivalent to the formal denition of the real numbers, and its proof requires several technical properties of convergent sequences. Using the Intermediate Value Theorem, we can prove that continuous functions must take on particular values, even if we cannot eplicitly say eactly where. Eample: Show that there is a real number c such that c = cos(c). If we let f() = cos(), then we want to show that f() attains the value zero somewhere. We can compute f(0) = and f(π) = π +. Since f() is continuous, and 0 lies between f(0) and f(), the Intermediate Value Theorem dictates that there is some c in (0, π) such that f(c) = 0. For this value of c, we have 0 = f(c) = c cos(c), so that c = cos(c) as desired. Eample: Show that the function p() = has at least three real zeroes. We will use the Intermediate Value Theorem on three separate intervals to show that there must eist a zero in each interval. If we make a small table of values, we will see that p( 2) = 06, p( ) = 3, p(0) = 6, p() =, and p(2) = 8. Notice that p() is continuous and changes sign in each of the intervals [ 2, ], [0, ], and [, 2]. Thus, applying the Intermediate Value Theorem on each interval shows that p() has at least one real zero in each of the intervals ( 2, ), (0, ), and (, 2), so it must have at least 3 zeroes in total. With all of these results, essentially the only remaining simple functions for which continuity is still an interesting question are piecewise-dened functions. For such functions, to determine continuity requires comparing one-sided its to the value of the function at the points where the denition of the function changes. 2 for < 2 Eample: Determine the values of a and b which make f() = + a for 2 3 continuous for all. b 2 for > 3 8

9 On the intervals (, 2), (2, 3), and (3, ), f is dened as a single polynomial, and is therefore continuous. The only possible issues are at = 2 and = 3, where the denition of f changes from one polynomial to another. As 2 from below, f() 4. As 2 from above, f() 2 + a, and f(2) = 2 + a. We need all three values to be equal, so 4 = a + 2 so that a = 2. Now as 3 from below, f() 3 + a = 5. As 3 from above, f() b 9, and f(3) = 3 + a = 5. We need all three values to be equal, so 5 = b 9 so that b = 4. Therefore, f() will be continuous when a = 2, b = 4. Here is a graph of y = f(), for a = 2 and b = 4:.5 Limits at Innity, Innite Limits Another type of behavior we are interested is how a function f() behaves when the value of becomes very large (either large positive or large negative). The precise denition (which, again, we will not use for evaluating its) is the following: We say that a function f() has the it L as +, written as f() = L, if the following + statement is true: for any ɛ > 0 (no matter how small) there eists an A > 0 (depending on ɛ) with the property that for all > A, we have that f() L < ɛ. We say that a function f() has the it L as, written as f() = L, if the following statement is true: for any ɛ > 0 (no matter how small) there eists an A > 0 (depending on ɛ) with the property that for all < A, we have that f() L < ɛ. Notation: The symbols and + mean the same thing (positive innity); the + is often used for contrast with the symbol (negative innity). All of the it laws also hold for its at innity, with all of the its being changed to the appropriate innite it. Note that some functions, such as f() =, do not tend to a nite value as becomes very large, but instead grow to innity. Other functions, such as g() = and h() = ln(), become very large even for nite 2 values of. So we will dene innite its to cover those cases: We say that a function f() diverges to + as c, written as c f() = +, if the following statement is true: for any B > 0 (no matter how large) there eists a δ > 0 (depending on B) with the property that for all 0 < c < δ, we have f() > B. The idea of diverging to is analogous, ecept instead f() < B. We can also formulate one-sided its here, with c from above or from below. 9

10 We say that a function f() diverges to + as, written as f() = +, if the following statement is true: for any B > 0 (no matter how large) there eists an A > 0 (depending on B) with the property that for all > A, we have f() > B. The statements for diverging to, or the statements as, are similar. There are some additional it laws that allow us to work with innite its (where a can be nite or innite): c = c for any constant c, + + Negation: If f() =, then [ f()] =, and vice versa. Basic Limits: = 0, =, and ln() =. 0+ Multiplication: If f() is a nite positive number or and g() =, then [f()g()] =. Addition: If f() is a nite number or and Division: If f() is a nite number and g() =, then g() =, then [f() + g()] =. [ ] f() = 0. g() Eponentiation ( L ): If f() = and g() = L where L > 0, or g() =, then is. Eponentiation (L ): If f() = L (where L 0) and g() =, then 0 L <, and is if L >. f() g() f() g() is 0 if Using these it laws for, we can attach a concrete meaning to certain arithmetic operations involving that allows us to treat and as if they were almost real numbers (subject to some restrictions): The negation statement says that = (i.e., we can move minus signs to and from in the way we would epect). The multiplication statement says and c are both for any positive constant c. The addition statement says that c + and + are both for any nite c. c The division statement says that = 0 for any nite c. The eponentiation statements say that a is 0 if 0 < a < and is if a >, and also that a = for a > 0. (Both statements include the fact that =.) By combining some of the above statements we can see what happens with as well: for eample, c = and ( ) ( ) =, and a is if a < and is 0 if a >. The other possible arithmetic operations (namely,, 0,, 0,, and 0 ) are indeterminate forms, meaning that their values depend on the functions whose its are involved. (They may be nite, innite, or not eist at all, depending on the situation.) Many innite its can be evaluated using direct appeals to the it laws and the basic its. Eample: Find n and n, for n a positive integer. By applying the eponentiation rule ( L ) for its with f() = and g() = n, we see that n =. For the other it, observe that n = ( ) n ( ) n. By applying the eponentiation rule ( L ) for its with f() = and g() = n (note that f() is positive when is large and negative), we see that ( )n =. { Thus, n = ( ) n if n is even ( )n = if n is odd. 0

11 Eample: Find + ( a) n,, and ( a) n, for n a positive integer. ( a) n By applying the eponentiation rule (L ) for its with f() = (which is positive when > a) a and g() = n, we see that + ( a) n =. For the other it, observe that ( a) n = ( ) n (a ) n. Applying the eponentiation rule (L ) for its with f() = (which is positive when < a) and g() = n, we see that a (a ) n =. Then ( a) n = ( )n = { if n is even if n is odd. Finally, by comparing the two one-sided its, we see Eample: Find 0. 2 { ( a) n = This eample is merely a special case of the previous one, with a = 0 and n = 2. From the analysis above, we see that the it eists and equals. Eample: Find e and e. if n is even DNE if n is odd. By applying the eponentiation rule for innite its we see e =. For the other it, we could either use the eponentiation rule again, or use the division rule to see that e = y e y = y Eample: Find ( 2) 6. = 0 (where we set y = in the rst step). ey Note that the it of the numerator as 2 is equal to, while the denominator goes to zero. From our earlier analysis, we know that 2 ( 2) 6 =. 2 5 Thus, by the multiplication law for innite its, we see that = ( ) ( ) =. 2 ( 2) 6 Other its require some algebraic manipulation or simplication to evaluate. In general, a good heuristic is that the behaviors of innite its tend to be determined by the largest terms. For its involving quotients of polynomials, or in general functions that can be compared to powers of, a useful idea is to factor out the highest power of that appears in the numerator and the denominator. Eample: Find ( ) and ( ). We should epect that the behavior of the it will be determined by the largest quantity, which in this case is 3. If we factor out the largest power of, we can write = 3 ( 2/ + 4/ 3 ). Taking the it as, we see that 3 while ( 2/ + 4/ 3 ). Thus, by the innite it laws for products, we see that ( ) =. Similarly, if we take the it as, we see that 3 while ( 2/ + 4/ 3 ). Thus, by the innite it laws for products, we see that ( ) =.

12 Eample: Evaluate 2 7. Here, 2 is the highest power that appears in both the numerator and denominator Factoring it out yields 2 7 = 2 [3 + 2/ / 2] 2 [ / 7/ 2 ]. Now we can use the it laws repeatedly to break down the evaluation into easy pieces, to see that [ 3 + 2/ / 2 ] [ = 3 and / 7/ 2 ] =, and so we obtain 2 7 = Eample: Evaluate 2 7. Like above, we pull out the highest power of that appears in the numerator and the denominator This gives 2 7 = 3 [3 + 2/ 2 / 3] 2 [ / 7/ 2 ] = 3 + 2/2 / 3 / 7/ / 2 / 3 We have = and = 3 by the same kind of argument as before, so by the / 7/2 multiplication rule for innite its we see that the original it was. Eample: Find and In each case, observe that the numerator and denominator both go to ±. To evaluate the it, we pull out the largest power of from each ( + 4/ For the rst it, = 2 ) 2 + 4/ = 2. 2 ( 2/) ( 2/) Since is large and positive, 2 + 4/ 2 =, so 2 + 4/ 2 = ( 2/) 2/ = = / For the second it, the same calculation yields = 2. 2 ( 2/) Since is large and negative, 2 + 4/ 2 =, so 2 + 4/ = 2 = ( 2/) 2/ =. [ Eample: Find ]. Here, there is a dierence involving a square root, so we use the idea of multiplying by the conjugate: [ ] = = ( 2 + 6) = Now we can pull out the largest power of from the numerator and denominator: = 6 + 6/ + = 6 = 6 + 6/ + + = 3. Well, you're at the end of my handout. Hope it was helpful. Copyright notice: This material is copyright Evan Dummit, You may not reproduce or distribute this material without my epress permission. 2