Proof of L'Hopital's RuleDate: 12/23/98 at 01:21:34 From: James Williams Subject: Proof of L'hopital's rule Can you show me a proof of L'Hopital's Rule? Thanks Date: 12/23/98 at 10:27:11 From: Doctor Anthony Subject: Re: Proof of L'hopital's rule To prove L'Hopital's Rule (sometimes spelled L'Hospital's Rule), we use the Taylor expansion: f(a+h) = f(a) + hf'(a) + terms in h^2 and higher g(a+h) = g(a) + hg'(a) + terms in h^2 and higher So: f(a+h) f(a)+h*f'(a) Lt ------ -> ------------ h->0 g(a+h) g(a)+h*g'(a) so with f(a) = g(a) = 0 we get: f(a+h) h*f'(a) f'(a) Lt ------- -> ------- -> ------ h->0 g(a+h) h*g'(a) g'(a) We can use l'Hopital's also if f'(a) -> infinity and g'(a) -> infinity: f(a) infinity 1/g(a) 0 ---- -> -------- so -------- -> --- g(a) infinity 1/f(a) 0 and applying l'Hopital's to this latter expression, we get: f(a) -g'(a)/[g(a)]^2 g'(a)*[f(a)]^2 ------ -> ---------------- -> ---------------- g(a) -f'(a)/[f(a)]^2 f'(a)*[g(a)]^2 and cross-multiplying: f'(a) f(a) ------- -> ------ g'(a) g(a) Therefore whether we have 0/0 or infinity/infinity we can use l'Hopital's rule. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ Date: 12/23/98 at 11:43:47 From: Doctor Rob Subject: Re: Proof of L'hopital's rule Thanks for writing to Ask Dr. Math! This is a rather complicated business. First of all, there are several forms of the Rule. I will state and prove one of them, and try to indicate how other forms are corollaries of this one. ====================================================================== Theorem (L'Ho^pital's Rule): Let f(x) and g(x) be differentiable on the interval a <= x < b, with g'(x) nonzero throughout. If: (i) lim f(x) = 0 and lim g(x) = 0 x->b- x->b- or if (ii) lim f(x) = infinity and lim g(x) = infinity x->b- x->b- and if: lim f'(x)/g'(x) = L x->b- then: lim f(x)/g(x) = L x->b- Note: b may be any real number or infinity, and L may be any real number or infinity. ====================================================================== Proof: We restrict our attention to the case where b and L are real numbers. Assume first hypothesis (i) is satisfied. We may set f(b) = 0 and g(b) = 0, and then f(x) and g(x) are continuous on the closed interval a <= x <= b. Applying Cauchy's Mean Value Theorem (see below), we see that for any x with a < x < b, there is a point x0 with x < x0 < b such that: f'(x0)/g'(x0) = [f(b)-f(x)]/[g(b)-g(x)] = f(x)/g(x) Since: lim f'(x0)/g'(x0) = L x0->b- given any epsilon > 0, we may choose delta > 0 so that: |f(x)/g(x) - L| = |f'(x0)/g'(x0) - L| < epsilon whenever b - delta(n) < x < b. This proves that: lim f(x)/g(x) = L x->b- This is the desired conclusion for hypothesis (i). - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Now assume hypothesis (ii). Given epsilon > 0, we first choose delta > 0 so that, whenever b - delta < x1 < b, then: |f'(x1)/g'(x1) - L| < epsilon Set x0 = b - delta, and take any points x with x0 < x < b. By Cauchy's Mean Value Theorem (see below), there is a choice of x1 with x0 < x1 < b, such that: [f(x)-f(x0)]/[g(x)-g(x0)] = f'(x1)/g'(x1) Thus: |[f(x)-f(x0)]/[g(x)-g(x0)] - L| < epsilon Now, defining: h(x) = [1-f(x0)/f(x)]/[1-g(x0)/g(x)] we can rewrite this as: |h(x)*f(x)/g(x) - L| < epsilon which is valid for all x with x0 < x < b. Now by using hypothesis (ii), it follows that: lim h(x) = 1 x->b- Choose x1 with x0 < x1 < b so that |h(x)-1| < epsilon and h(x) > 1/2, whenever x1 < x < b. For such values of x, we have: |h(x)*[f(x)/g(x)-L]| = |h(x)*f(x)/g(x) - L*h(x)| <= |h(x)*f(x)/g(x) - L| + |L*[1-h(x)]| < epsilon + |L|*epsilon and: |f(x)/g(x) - L| < (1+|L|)*epsilon/h(x) < 2*(1+|L|)*epsilon This proves that: lim f(x)/g(x) = L x->b- Q.E.D. ====================================================================== Now the proof above depends in two places on the following. Cauchy's Mean Value Theorem: Suppose that two functions f(x) and g(x) are continuous in the closed interval a <= x <= b and differentiable in the open interval a < x < b. Suppose further that g'(x) is nonzero for x in a < x < b. Then there exists at least one number c with a < c < b such that: [f(b)-f(a)]/[g(b)-g(a)] = f'(c)/g'(c) ---------------------------------------------------------------------- Proof: Let: h(x) = [g(b)-g(a)]*[f(x)-f(a)] - [f(b)-f(a)]*[g(x)-g(a)] Then clearly h(a) = h(b) = 0. Since h is continuous in [a,b] and differentiable in (a,b), we can apply Rolle's Theorem, which tells us that there exists at least one number c in (a,b) such that h'(c) = 0. Then: h'(c) = [g(b)-g(a)]*f'(c) - [f(b)-f(a)]*g'(c) = 0 Rearranging this equation, and using the fact that g'(c) is nonzero, we get: [f(b)-f(a)]/[g(b)-g(a)] = f'(c)/g'(c) Q.E.D. ====================================================================== Now let's talk about variations on L'Ho^pital's Rule. The first is to see that if L = +infinity or -infinity, the same ideas work to give a proof, although we have to modify the parts dealing with epsilons and deltas to accommodate the definition of an infinite limit. Instead of |F(x) - L| < epsilon, you need |F(x)| > 1/epsilon, and so on. The second is to see that if we replace the limits with x->b- with ones with x->b+, and replace the condition a <= x < b with b < x <= a, the conclusion still holds. This is a corollary of the stated theorem, obtained by substituting y = 2*b - x, and applying the theorem. The third is to see that if we replace the one-sided limits by two- sided ones, the conclusion still holds, as a corollary of the theorem and the immediately preceding paragraph. The fourth is to see that if we take b = +infinity or -infinity, the conclusion of the theorem still holds. We prove this by substituting x = 1/y, and then as x -> +infinity, y -> 0+, and as x -> -infinity, y -> 0-. Also, dx/dy = -1/y^2, and by the chain rule: f'(x) = d/dy[f(1/y)] = f'(1/y)*d/dy[1/y] = f'(1/y)*(-1/y^2) and similarly for g'(x). Thus: lim f'(x)/g'(x) = lim [f'(1/y)*(-1/y^2)]/[g'(1/y)*(-1/y^2)] x->+infinity y->0+ = lim f'(1/y)/g'(1/y) y->0+ = lim f(1/y)/g(1/y) y->0+ by the theorem, so: lim f'(x)/g'(x) = lim f(x)/g(x) x->+infinity x->+infinity which was to be shown. The case x -> -infinity is handled in the same way. I know this is long and complicated, but at least I warned you! - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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