Chapter Differentiation Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 6.3, Slide Section 6.3 L’Hospital’s Rule Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 6.3, Slide lim Consider the limit x→c f ( x) g ( x) where f (x) → L and g(x) → M If g(x) ≠ for x close to c and M ≠ 0, then Theorem 5.1.13 says that lim x→c f ( x) L = g ( x) M 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 lim x→0 Then f ( x) kx = lim = lim k = k x→0 g ( x) x → x So the indeterminate form (sometimes labeled 0/0) can lead to any real number as the limit Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 6.3, Slide Theorem 6.3.1 (Cauchy Mean Value Theorem) Let f and 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) 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) a) = h(b) Furthermore, Thus, by the mean value theorem (or Rolle’s Theorem), there exists c ∈ (a, b) such that h′(c) = That is, [ f (b) – f (a)] g ′(c) – [g (b) – g (a)] f ′(c) = ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 6.3, Slide In the expression [ f (b) – f (a)] g ′(c) = [g (b) – g (a)] f ′(c), f (b) − f (a ) f ′(c) = , g (b) − g (a ) g ′(c) or, f (b) − f (a ) = f ′(c) b−a g(x) = x for all x, then we have This formula is the original mean value theorem Theorem 6.3.2 (L’Hospital’s Rule) 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) = Suppose also that g ′(x) ≠ for x ∈ U, where U is the intersection of (a, b) and some deleted neighborhood of c If lim x→c f ′( x) =withL,L ∈ ′ g ( x) Copyright © 2013, 2005, 2001 Pearson Education, Inc , then lim x→c f ( x) = L g ( x) Section 6.3, Slide f ′( x) lim = L x → c g ′( x ) We have c ∈ [a, b], f (c) = g (c) = 0, g ′(x) ≠ for x ∈ U, and Let (xn) be a sequence in U that converges to c Proof: Apply the Cauchy mean value theorem to f and 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) ≠ for all x ∈ U, and g (c) = 0, we must have g (xn) ≠ for all n (See Rolle’s Theorem.) Thus, since f (c) = g(c) = 0, we have L L f ( xn ) f ( xn ) − f (c) f ′(cn ) = = g ( xn ) g ( xn ) − g (c ) g ′(cn ) for all n Furthermore, since xn → c and cn is between xn and c, it follows that cn → c Thus by Theorem 5.1.8, limso x→c lim n→∞ f ′(cn ) = L ′ g (cn ) But then, lim n→∞ f ( xn ) = L, g ( xn ) f ( x) = L , also ♦ g ( x) Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 6.3, Slide Example 6.3.3 Then f (1) = g(1) = Let f (x) = 2x – 3x + and g(x) = x – And we have f ′(x) = 4x – and g ′(x) = 1, so that x − 3x + 4x − lim = lim =1 x →1 x →1 x −1 We did this same problem by factoring in Example 5.1.5 Example 6.3.4 Then f (0) = g(0) = et f (x) = – cos x and g(x) = x And we have f ′(x) = sin x and g ′(x) = 2x, so that − cos x sin x cos x = lim = lim = x→0 x → 2x x→0 2 x2 lim provided that the second limit exists Since sin x → and 2x → as x → 0, 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) = Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 6.3, Slide 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 L is the limit of f as x → ∞, and we lim f ( x) = write L, provided that for each ε > there exists a real number N > a x→∞ such that x > N implies that | f (x) – L | < ε 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, ∞) → lim f ( x) = ∞ We say that f tends to ∞ as x → ∞, and we write provided that given any α ∈ x→∞ there exists an N > a such that x > N implies that f (x) > α Using these definitions, we can state l’Hospital’s rule for indeterminates of the form ∞/∞ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 6.3, Slide Theorem 6.3.8 (L’Hospital’s Rule) Let f and g be differentiable on (a, ∞) Suppose that limx → ∞ f (x) = limx → ∞ g (x) = ∞, and that g ′(x) ≠ for x ∈ (a, ∞) If lim x→∞ f ′( x) = L, g ′( x)with L ∈ , then lim x→∞ f ( x) = L g ( x) 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 lim x → c f (x) = lim x → c g(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 ⋅ ∞, , , ∞ , and ∞ – ∞, and are evaluated by using algebraic manipulations, logarithms, or exponentials to change them into one of the forms 0/0 or ∞/∞ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 6.3, Slide Example 6.3.10 For x > 0, let f (x) = x and g(x) = – ln x Then lim x → 0+ f (x)g(x) is an indeterminate of the form ⋅ ∞ To evaluate the limit, we write ∞ − ln x x → 0+ / x lim ( x)( − ln x) = lim x → 0+ = lim x → 0+ −1/ x −1/ x = lim x = x → + ∞ From this we also conclude that lim x → 0+ x ln x = Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 6.3, Slide 10 ... for x ∈ U, and Let (xn) be a sequence in U that converges to c Proof: Apply the Cauchy mean value theorem to f and g on the intervals [xn, c] or [c, xn] to obtain a sequence (cn) with cn between... (x) → L and g(x) → M If g(x) ≠ for x close to c and M ≠ 0, then Theorem 5.1.13 says that lim x→c f ( x) L = g ( x) M If both L and M are zero, we can sometimes evaluate the limit by canceling... two functions that can give rise to ambiguous values These indeterminate forms are indicated by the symbols ∞ 0 ⋅ ∞, , , ∞ , and ∞ – ∞, and are evaluated by using algebraic manipulations, logarithms,