Chapter 6 Differentiation
Section 6.3 L’Hospital’s Rule
Consider the limit where f(x) L and g(x) M. If g(x) 0 for x close to c and M 0, then Theorem 5.1.13 says that If both L and M are zero, we can sometimes evaluate the limit by canceling a common factor in the quotient. In this section we derive another technique that is often easier to use than factoring and has wider application. In general, when L= M = 0, the limit of the quotient f /g is called an indeterminate form, because different values may be obtained for the limit. For example, for any real number k, let f(x) = kx and g(x) = x. Then So the indeterminate form (sometimes labeled 0/0) can lead to any real number as the limit.
Theorem 6.3.1 (Cauchy Mean Value Theorem) Letfand g be functions that are continuous on [a,b] and differentiable on (a,b). Then there exists at least one point c (a,b) such that [f(b) –f(a)]g(c)=[g(b) – g(a)]f(c). Proof: Let h(x)=[f(b)– f(a)]g(x) – [g(b)– g(a)] f(x) for each x[a,b]. Then h is continuous on [a,b] and differentiable on (a,b). Furthermore, h(a) =[f(b)– f(a)]g(a) – [g(b)– g(a)] f(a) = f(b)g(a) – g(b)f(a) h(b)= [f(b)– f(a)]g(b) – [g(b)– g(a)] f(b) = f(b)g(a) – g(b)f(a) Thus, by the mean value theorem (or Rolle’s Theorem), there exists c (a,b) such that h(c) = 0. So, h(a) = h(b). That is, [f(b) –f(a)]g(c) – [g(b) – g(a)]f(c)=0.
In the expression [f(b) –f(a)]g(c)=[g(b) – g(a)]f(c), or, if g(x) = x for all x, then we have This formula is the original mean value theorem. Theorem 6.3.2 (L’Hospital’sRule) Let f and g be continuous on [a,b] and differentiable on (a,b). Suppose that c [a,b] and that f(c) = g(c) = 0. Suppose also that g(x) 0 for xU, where U is the intersection of (a,b) and some deleted neighborhood of c. If with L , then
We have c [a,b], f(c) = g(c) = 0, g(x) 0 for xU, and Let (xn) be a sequence in Uthat converges to c. Proof: Apply the Cauchy mean value theorem tofand g on the intervals [xn, c] or [c, xn] to obtain a sequence (cn) with cn between xn and c for each n, such that [f(xn) –f(c)]g(cn)=[g(xn)–g(c)]f(cn). Since g(x) 0 for all xU, and g(c) = 0, we must have g(xn) 0 for all n. (See Rolle’s Theorem.) Thus, since f(c) = g(c) = 0, we have L L for all n. Furthermore, since xnc and cn is between xn and c, it follows that cnc. Thus by Theorem 5.1.8, But then, so , also.
Example 6.3.3 Then f(1) = g(1) = 0. Letf(x) = 2x2 – 3x + 1 and g(x) = x – 1. And we have f(x) = 4x – 3 and g(x) = 1, so that We did this same problem by factoring in Example 5.1.5. Example 6.3.4 Letf(x) = 1 – cosx and g(x) = x2. Then f(0) = g(0) = 0. And we have f(x) = sin x and g(x) = 2x, so that provided that the second limit exists. Since sin x0 and 2x0 as x0, we again have the indeterminate form 0/0. So we use l’Hospital’s rule again. Note that g(x) must be nonzero in a deleted neighborhood of 0, but it is permitted that g(0) = 0.
In some situations we wish to evaluate the limit of a function for larger and larger values of the variable. Definition 6.3.6 Let f: (a, ) . We say that the real number Lis the limit of fas x, and we write provided that for each > 0 there exists a real number N>a such that x>N implies that |f(x) – L| <. Very often as x the values of a given function also get large. This leads to the following definition. Definition 6.3.7 Let f: (a, ) . We say thatftends to as x, and we write provided that given any there exists an N>a such thatx>N implies thatf(x) >. Using these definitions, we can state l’Hospital’s rule for indeterminates of the form /.
(L’Hospital’s Rule) Let f and g be differentiable on (a,). Suppose that limx f(x)=limxg(x)=, and that g(x) 0 for x (a,). If with L , then Theorem 6.3.8 The proof is lengthy, but basically consists of the Cauchy mean value theorem and algebra. There is also an extension of l’Hospital’s rule that applies to f/g when limx cf (x) = limx cg(x) = and c . (See Exercise 13.) There are other limiting situations involving two functions that can give rise to ambiguous values. These indeterminate forms are indicated by the symbols 0, 00, 1, 0, and – , and are evaluated by using algebraic manipulations, logarithms, or exponentials to change them into one of the forms 0/0 or/.
Example 6.3.10 For x > 0, let f (x) = x and g(x) = – lnx. Then limx 0+ f (x)g(x) is an indeterminate of the form 0 . To evaluate the limit, we write = 0. From this we also conclude that limx 0+ xlnx = 0.