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L'Hospital's rule 4.26. Prove L'Hospital's rule for the case of the "indeterminate forms" (a) 0/0 and (b) 0/00. (a) We shall suppose that f(x) and g(x) are differentiable in a

L'Hospital's rule 4.26. Prove L'Hospital's rule for the case of the "indeterminate forms" (a) 0/0 and (b) 0/00. (a) We shall suppose that f(x) and g(x) are differentiable in a

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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4.26) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text.
L'Hospital's rule 4.26. Prove L'Hospital's rule for the case of the "indeterminate forms" (a) 0/0 and (b) ∞/0. (a) We shall suppose that f(x) and g(x) are differentiable in a <x<b and f(x,) = 0, g(x,) = 0, where a<x,<b. By Cauchy's generalized mean value theorem (Problem 4.25), f(x) _ f(x)- f(x,) _ f'CE) g(x)- g(x,) g'E) X。くらくx g(x) Then f(x) lim f'(x) f'(E) = lim = lim = L x→Xo+ g(x) x→X,+ g'(E) g'(x) since as x → xo+, Š → xo+. Modification of this procedure can be used to establish the result if x → xo -, x → Xo, X → 0, or x→-0. (b) We suppose that f(x) and g(x) are differentiable in a <x< b, and lim f(x) = ∞, lim g(x)=∞ where x→x,+ x→x,+ a<xo<b.
Transcribed Image Text:L'Hospital's rule 4.26. Prove L'Hospital's rule for the case of the "indeterminate forms" (a) 0/0 and (b) ∞/0. (a) We shall suppose that f(x) and g(x) are differentiable in a <x<b and f(x,) = 0, g(x,) = 0, where a<x,<b. By Cauchy's generalized mean value theorem (Problem 4.25), f(x) _ f(x)- f(x,) _ f'CE) g(x)- g(x,) g'E) X。くらくx g(x) Then f(x) lim f'(x) f'(E) = lim = lim = L x→Xo+ g(x) x→X,+ g'(E) g'(x) since as x → xo+, Š → xo+. Modification of this procedure can be used to establish the result if x → xo -, x → Xo, X → 0, or x→-0. (b) We suppose that f(x) and g(x) are differentiable in a <x< b, and lim f(x) = ∞, lim g(x)=∞ where x→x,+ x→x,+ a<xo<b.
Assume x, is such that a < xo <x<x¡ <b. By Cauchy's generalized mean value theorem, f(x)- f(x,) _ f'() x< < x, %3D g(x)- g(x,) g'(5 ) Hence, f(x)- f(x,) _ f(x) 1- f(x,)/ f(x) f'C) g(x)- g(x,) g(x) 1- g(x,)/g(x) g'(5) from which we see that f(x) _ f'(E) 1-g(x,)/g(x) g(x) g'() 1-f(x,)/f(x) (1) f'(x) x>Xo + g'(x) Let us now suppose that lim = L and write Equation (1) as f(x) ( f'(). 1- g(x, )/g(x) 1-8(x, Vg(x) - L + L (2) 8(x) g'(x) 1- f(x, )/f(x) We can choose x, so close to x, that |f'(5)/g'(E) - L| < e. Keeping x, fixed, we see that 1-8(x,)/g(x) lim |=1 since 1 lim f(x), = ∞ and lim g(x) = 0 1-f(x,)/f(x) xXo + xXo + x-Xo + Then taking the limit as x → xo+ on both sides of (2), we see that, as required, f(x) f'(x) = L = lim lim x→Xq+ g(x) x->Xo+ g'(x) Appropriate modifications of this procedure establish the result if x → xo -, x → Xo, x →∞, or x → -.
Transcribed Image Text:Assume x, is such that a < xo <x<x¡ <b. By Cauchy's generalized mean value theorem, f(x)- f(x,) _ f'() x< < x, %3D g(x)- g(x,) g'(5 ) Hence, f(x)- f(x,) _ f(x) 1- f(x,)/ f(x) f'C) g(x)- g(x,) g(x) 1- g(x,)/g(x) g'(5) from which we see that f(x) _ f'(E) 1-g(x,)/g(x) g(x) g'() 1-f(x,)/f(x) (1) f'(x) x>Xo + g'(x) Let us now suppose that lim = L and write Equation (1) as f(x) ( f'(). 1- g(x, )/g(x) 1-8(x, Vg(x) - L + L (2) 8(x) g'(x) 1- f(x, )/f(x) We can choose x, so close to x, that |f'(5)/g'(E) - L| < e. Keeping x, fixed, we see that 1-8(x,)/g(x) lim |=1 since 1 lim f(x), = ∞ and lim g(x) = 0 1-f(x,)/f(x) xXo + xXo + x-Xo + Then taking the limit as x → xo+ on both sides of (2), we see that, as required, f(x) f'(x) = L = lim lim x→Xq+ g(x) x->Xo+ g'(x) Appropriate modifications of this procedure establish the result if x → xo -, x → Xo, x →∞, or x → -.
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