Limits and Derivatives (Review of Math 249 or 251)

Transcription

1 Chapter 3 Limits and Derivatives (Review of Math 249 or 251) 3.1 Overview This is the first of two chapters reviewing material from calculus; its and derivatives are discussed in this chapter, and integrals will be discussed in the net. While formal definitions of its and derivatives are included, for the sake of completeness, these won t be used very much in CPSC 413; you should concentrate more on the rules for computing its and derivatives that this chapter includes. 3.2 Limits Informal Introduction You should remember from Math 249 or 251 that its reveal the behaviour of a function near a point. Consider, for eample, the function f() = A graph of the function near = 1, and the function values f(0.9)=1.81, f(0.99)=1.9801, f(0.999) = , as well as f(1.1)=2.21, f(1.01)=2.0201, and f(1.001) = , should both suggest that the it of the value of this function as approaches 1 is 2. That is, these should suggest that f() =2. 1 Of course, this is f(1) in this case, so that for this eample the it of the value of the function as approaches 1 is just f(1). This isn t always true. For eample, If f() = 1 there is no it of the value f() asapproaches 0: The value diverges to + as approaches 0 from the right (that is, as takes on progressively smaller positive values), and it diverges to as approaches 0 from the left (that is, as takes on progressively smaller negative values). If f() = the it of f() does not eist as approaches 0. This time, though, the value approaches 1 from the right and 1 from the left (and, again f(0) is not defined). 41

2 If f() = 2, the it of f() eists as approaches 0, and the it equals 1, but f(0) is not defined. In fact, f() = 1 for all ecept when =0. Finally, if { 1 if 0 f() = 0 if =0, the it of f() asapproaches 0 eists, and f(0) eists, but these values aren t the same: The it is 1 and the function value is Formal Definition Note: This section is, pretty much, for interest only. That is, it won t really be necessary to understand or use the formal definition of a it in CPSC 413. Informally, we say that f() =Lif f() gets arbitrarily close to L as approaches a. That is, if you give me any positive real number ɛ that defines a neighbourhood (that is, open interval) (L ɛ, L + ɛ) near the it L, I should be able to give back to you another positive real number δ defining a neighbourhood (a δ, a + δ) near the value is approaching, so that f() will lie in your neighbourhood whenever falls into my neighbourhood as long as is different from a. A more formal definition, which pretty much says the same thing, is as follows. Definition 3.1. The it of f() as approaches a eists and is equal to (a finite constant) L, f() =L, if, for every real number ɛ>0, there eists some real number δ>0 such that if 0 < a <δ f() L <ɛ. (3.1) Note (again) that this does not imply that f(a) =L, or even that f(a) is defined, because the definition only mentions a condition that must hold when a is strictly greater than zero. In CPSC 413 we ll frequently be interested in the its of function values as becomes arbitrarily large (as approaches + ): Definition 3.1, Continued: The it of f() asapproaches + eists and is equal to (a finite constant) L, f() =L, + if, for every real number ɛ>0, there eists some (generally, large) real number M>0 such that if >M f() L <ɛ. Finally, the it of f() asapproaches + eists and is equal to +, f() =+ + if, for every real number U>0, there eists some (generally, large) real number M>0 such that if >M f() >U. 42

3 If all else fails, it is occasionally possible to prove that f() =Lusing this definition. Suppose again, for eample, that f() = and let a = 1. To prove formally (from the definition) that f() =Lin this case, for L = 2, suppose that is a small real number, and note that f(1+ )=(1+ ) 2 +1= It is sufficient to consider small values for, so suppose now that < 1. Then 2 < and it follows from the above that in this case. Therefore 2 2 <f(1+ )<2+3 f(1+ ) 2 <3. Now, given any positive real number ɛ, we can choose δ to the the minimum of 1 and ɛ/3. Then δ is also a positive real number, and it follows from the above derivation that whenever f() 2 <ɛ 1 <δ, establishing (formally) that 1 f() = 2, as desired. A formal proof that a it eists, and and that it equals some specified value, generally has the above form: You must use algebraic manipulation to figure out what a specific value for δ is, for any given value of ɛ, so that inequality 3.1 holds Limits of Functions Defined on the Integers The above definition describes the it of a function that s defined on the real numbers. However, in CPSC 413, we ll frequently want to compute a it, as the evaluation point approaches infinity, of a function that s only defined on the integers. In order to apply the above definition and the following evaluation rules we ll just etend the definition of the function we re interested in. Sometimes it ll be obvious how to do this, because the description of the function that s only defined on the integers gives a way to define it at other real values, too. For eample, if you want to compute the it of the identity function f(n) =nit should be clear that you should etend this by defining f() tobefor every real number as well. Unfortunately, this isn t always the case. For eample, it isn t obvious how to etend the definition of the factorial function f(n) =n! in order to define it on real numbers that aren t integers. If all else fails, the following trick will work: If we want to compute f(n), where f(n) is n + defined on the integers, and it isn t clear how to define a value for f at other real numbers, we ll define a corresponding function ˆf that s defined on the reals by the rule ˆf() =f( ) for every real number, 43

4 where, as usual, is the largest integer that s less than or equal to (and is called the floor of ). Note that f(n) = ˆf(n) for every integer n according to this definition. Now we ll just define f(n) tobethe same as ˆf(), so that the evaluation rules can n + + be applied to the second of these its, in order to find the value for the first one. You might be concerned because there are several different ways to etend a function that s defined on the integers, to get a function that s defined on the reals. If any of the etensions has a it, this really will be the it (as n approaches infinity) of the integer function you started with and all of the other possible etensions will either have the same it or won t have any it at all. Furthermore, the etension suggested above (involving the use of as an argument) will have a it if any of the etensions do. Another way to deal with this problem will be given as part of Theorem 3.6, below. Finally, please note that the above tricks are only recommended when you want to compute a it at infinity don t apply them if you want to compute its as the evaluation point approaches a finite value! Rules for Calculating Limits The following rules aren t quite as general as the use of the definition of it, but they re easier to apply and will be sufficient for CPSC 413 (at least, most of the time): Theorem 3.2 (Limit of a Constant). If f() =c(that is, f is a constant function), f() =c for any real or comple number a, orfora=+. Theorem 3.3 (Sum, Difference, and Product Rules). Suppose that f() =L and g() =M for functions f and g, finite values L and M, and for any real or comple number a (or for a =+ ). Then (f()+g()) = L + M, and (f() g()) = L M, (f() g()) = L M. Theorem 3.4 (Quotient Rule). If a, f(), g(), L and M are as above, and M 0, f() g() = L M. 44

5 You can etend these rules to some of the cases where one or both of L and M equal plus or minus infinity. Unfortunately, there are some cases where you can t: The sum and the quotients (+ )+( ) 0 0 and + + are all ill-defined it isn t clear what values they should assume. The last of these (the ratio of + to itself) will arise frequently in CPSC 413, so we need some additional ways to deal with it. Theorem 3.5 (Cancellation of a Common Factor). Let f, g, and h be functions, let a be a finite constant, and suppose there is some positive real number δ such that if 0 < a <δ h() 0; f() h() g() h() = f() g(). Similarly, if there is some (generally large) real number M such that if >M h() 0, f() h() + g() h() = f() + g(). That is, you can cancel out the common factor of h() in the numerator and denominator without changing the it in each case. Sometimes this is all you need to do in order to go ahead and compute the it. For eample, consider the it Let h() =; h() is nonzero when is close to 0, in the sense given above, even though h(0) = 0, and the it has the form described above for f() =3andg() = 2. So we can use Theorem 3.5 to conclude that = = 3 2. Here s another fact that is occasionally useful. This doesn t only apply to its of ratios, but it can definitely be useful in this case. Theorem 3.6 (Sandwiching the Limit). Let f, g, and h be functions that are defined on the real numbers such that, for a finite constant a and positive real number δ, if 0 < a <δ f() g() h(). 45

6 Suppose, furthermore, that f() = h(); g() = f() as well. Similarly, if f, g, and h are functions that are defined on the real numbers and N is a (generally large) constant such that and if >N f() g() h() f() = h() + + g() = f() + + as well. Finally, if f and h are functions that are defined on the reals, g is a function that is defined on the integers, and N is a (generally large) constant such that (for integers n and real numbers ) and as well. if n>n f(n) g(n) h(n) f() = h() + + g(n) = f() n + + This result can be applied to compute the middle it for g as long as f and h really do provide lower and upper bounds on g within the range you re interested in (that is, near the it point), and as long as the its of f and h really do agree. Here s an eample: Suppose you want to compute the it 2 n n + n! It isn t clear how you can do this directly. On the other hand, since n n! = i, i=1 46

7 it isn t too hard to see that if n 1 so that Set 2 9 3n =2 3 n 2 =1 2 n n n i = n! n = n n, i=3 i=3 i=1 2 n n n 2n n! 9 2 2n 3 n = 9 ( ) 2 n 2. (3.2) 3 h() = 2 9 ( ) 2 ; 3 clearly h is defined on the real numbers, and ( ) h() = =0, since < 1. We could try to set f() tobe 2 (based on the inequality given above) and apply the method being described. However, it ll make things easier if we simply observe that 2 is nonnegative for all 0, and use the inequality 0 2n n! 9 ( ) 2 n 2 3 for every natural number n that is implied by this observation and by inequality (3.2), above. Therefore we ll set f() =0,g(n)= 2n n!, and h() to be as defined above, and we ll set N to be 1. Then and clearly (since f() =0) f(n) g(n) h(n) for all n N, so we can conclude that f() = h() =0, + + g(n) =0 n 0 as well. Note that this is the it that we originally wished to compute. We ll see one more etremely useful rule to compute the its of ratios, after defining derivatives Continuous Functions Definition 3.7. A function f() iscontinuous at a if the it as approaches a eists and, furthermore, f() =f(a). A function is continuous on an interval if it s continuous at every point inside that interval, and it s continuous if it s continuous at every real number. 47

8 By definition, the its of continuous functions are etremely easy to compute: the it of a continuous function is just the value of the function at the it point. Polynomials and eponential functions are continuous, and log functions (f() = log c for some constant c>0) are continuous on the interval of positive real numbers. You may use these facts without having to prove them in CPSC 413. Of course, this still leaves us with the problem of computing its as approaches Derivatives Definitions Suppose f() is a function and a is a real number such that f is continuous at a. Informally (or graphically), you can think of f (a) as the slope of the line that is tangent to the graph y = f() at the point (a, f(a)). If f is not continuous at a f (a) doesn t eist. The derivative of f at a can be defined more formally as a it: Definition 3.8. If a function f is continuous at a point a the derivative of f at a, f (a), is the it δ 0 f(a + δ) f(a). δ The function f is said to be differentiable at a if f is continuous at a and the above it eists (otherwise, f (a) is undefined), and f is said to be differentiable if f (a) eists for every real number a. If f is differentiable we ll frequently consider f to be a function as well that is, we ll refer to the function f () (or just f ), with the understanding that the value of this function at any real number a is the it f (a) given above Some Rules for Computing Derivatives Theorem 3.9 (Derivative of a Constant). If f() =cis a constant function f () =0. Theorem 3.10 (Derivative of a Power). If f() = α, for any fied real number α, f is differentiable and f () =α α 1. In particular, the derivative of 1 (the function 0 )is0. Theorem 3.11 (Sum, Product, and Quotient Rules). If functions f and g are both differentiable at a, if h = f + g, h (a) =f (a)+g (a); and if h = f g, h (a) =f(a) g (a)+f (a) g(a); if h = f g and g(a) 0, h (a) = f (a) g(a) f(a) g (a) (g(a)) 2. 48

9 Theorem 3.9 and the above sum and product rules can be used to establish that differentiation is a linear operator: If c is a constant and f, g, and h are functions such that f() =c g()+h(), f () =c g ()+h () as well. Theorem 3.12 (Chain Rule). If g is differentiable at a, f is differentiable at g(a), and if h = f g (so that h() =f(g())), h (a) =f (b) g (a), for b = g(a). 3.4 Back to Limits: l Hôpital s Rule Now that differentiation has been defined, we can state another etremely useful rule for computing its of quotients. Theorem 3.13 (l Hôpital s Rule). If a is a real number or a =+, and either or f() = g() =0 f() = g() =+, f() g() = f () g () (provided, of course, that the derivatives shown in the above epression eist). That is, you can differentiate the functions in the numerator and in the denominator, without changing the it, in these cases. The first eample of the use of l Hôpital s rule you saw might have been the following: Suppose f() = sin and g() =; f() = g() = 0. So, by l Hôpital s rule, 0 0 sin 0 = cos =cos0=1, 0 1 since f () = cos if f() = sin and g () =1ifg()=, and since the cosine function is continuous. Note, again, that you can only apply this rule if either f() and g() both approach 0 or if they both approach + at the it you want to compute. On the other hand, you won t need this rule in any other case. 3.5 Eponential and Logarithmic Functions In spite of the fact that the last eample involved one, trigonometric functions (sin, cos, tan, and so on) won t be used very much on CPSC 413. However, the eponential and logarithmic functions will be used frequently, and you should know and be able to apply the basic properties of these functions that are given below. Here are definitions and derivatives for two important functions (and a constant). 49

10 Definition Euler s constant, commonly written as e, is the it ( 1+ 1 ) n n + n and is approimately equal to Theorem If f() =e f () =e as well. Definition If >0 the natural logarithm of, ln, is the (unique!) real number y such that e y =. Thusifg()=lnand f() =e as above, for all >0. f(g()) = g(f()) = Theorem If g() =lnas above, and >0, g () = 1. We re often interested in eponential and logarithm functions with different bases. You can differentiate these as well, by using the following formulas and the above definitions. Theorem If c>0 c = e ln c, Theorem If a, b, c > 0 In particular (when a is Euler s constant), log b c = log a c log a b. log b c = ln c ln b. You should now be able to use the fact that if f() =e f () =e as well, in order to figure out how to differentiate c for c>0 (by using the above formulas and the chain rule). It should be even easier to figure out how to differentiate log b for some constant b>0, given the above information about logarithms (after all, if b is a positive constant, ln b and 1 ln b are just constants, as well). Some additional useful properties of eponential functions and logarithms are given on pages of Cormen, Leiserson, and Rivest [3]. 3.6 Eercises 1. Compute the each of the following its or eplain why it does not eist. 3 8 (a) 2 2 (b) + (c) cos 1 (d) 0 sin(4) sin(3) 50

11 2. Compute the derivative (with respect to ) of each of the following functions. (a) f() = (b) f() =ln (c) f() = ln (d) f() =e 2 ln 3. Derive the quotient rule (as given below), starting with the definition of a derivative, and assuming that the functions f and g are both differentiable at a and that g(a) 0. Quotient Rule: If h() = f() g() h (a) = f (a)g(a) f(a)g (a) (g(a)) 2 4. Prove that for every natural number n 1. (ln ) n + =0 Hints for selected eercises are given in the net section; solutions for Eercises 1(c) and 3 are given in Section Hints for Selected Eercises Eercise #1(a): Observe that you can cancel out a common factor of 2 from both the numerator and the denominator of the epression in the it or you can use l Hôpital s rule, instead. (Try using both methods, and confirm that you get the same answer each time.) Eercise #1(b): You can either treat 2 as a common factor to be einated from both the numerator and denominator, or you can apply l Hôpital s rule twice. Eercise #1(d): Apply l Hôpital s rule. You ll need to treat both the numerator and denominator as compositions of functions and use the chain rule to compute the needed derivatives. Eercise #2(b): Eercise #2(c): Apply the product rule for derivatives. Apply the quotient rule for derivatives. Eercise #2(d): f() =g(h()) for g(y) =e y and h(z) =z 2 ln z. Apply the product rule to differentiate h, and apply the chain rule to differentiate f. Eercise #4: Try to prove this using mathematical induction, by induction on n (with n = 1as your starting point). You ll probably need to apply l Hôpital s rule in both the basis and inductive step of your proof. 51