Limit Proofs with L'Hopital's RuleDate: 05/27/98 at 10:28:31 From: Kumarpal Sheth Subject: Proofs for limits Hello Dr. Math, I have been just introduced to calculus. In limits, we have the following identities as results without proof: 1) lim (x + 1/x)^x = e x-->0 2) lim (a^x - 1)/x = ln(a) x-->0 3) lim (ln(1+x))/x = 1 x-->0 Date: 05/27/98 at 13:26:12 From: Doctor Rob Subject: Re: Proofs for limits Later on in your study of calculus you will learn something called L'Ho^pital's Rule. This will allow you to compute and prove these limits. It says: Theorem: If as x -> a (where a is any real number or infinity), lim f(x) = 0 and lim g(x) = 0, then: lim [f(x)/g(x)] = lim [f'(x)/g'(x)], x->a x->a provided either limit exists (in which case both do). Corollary: If lim f(x) = infinity and lim g(x) = infinity, then: lim [f(x)/g(x)] = lim [f'(x)/g'(x)], x->a x->a provided either limit exists (in which case both do). Proof of Corollary: We start with: lim f(x)/g(x) = lim [1/g(x)]/[1/f(x)] x->a x->a and we can apply the Theorem to this quotient: lim f(x)/g(x) = lim [-g'(x)/g(x)^2]/(-f'(x)/f(x)^2] x->a x->a = {lim [g'(x)]/f'(x)]}*{lim [f(x)/g(x)]}^2 x->a x->a 1/lim [g'(x)/f'(x)] = lim [f(x)/g(x)] x->a x->a lim [f(x)/g(x)] = lim [f'(x)/g'(x)] Q.E.D. x->a x->a If you accept this theorem without proof, the above three limits can be computed using it: 1) lim (x + 1/x)^x = lim e^(x * ln[x + 1/x]) x->0 x->0 = e^lim ln[x + 1/x]/(1/x) x->0 Now the limit has the form lim f(x)/g(x), where f(x) = ln(x + 1/x) and g(x) = 1/x, and lim f(x) = infinity, and lim g(x) = infinity as x -> 0. Apply the corollary to L'Ho^pital's Rule: f'(x) = (x^2 - 1)/[x * (x^2 + 1)] g'(x) = -1/x^2 f'(x)/g'(x) = -x * (x^2 - 1)/(x^2 + 1) The limit of this as x -> 0 is 0. Thus: lim (x + 1/x)^x = e^lim -x*(x^2-1)/(x^2+1) x->0 x->0 = e^0 = 1 2) lim (a^x - 1)/x x->0 This is already in the correct form for L'Ho^pital's Rule, with f(x) = a^x - 1 and g(x) = x. The hard part here is computing f'(x), but a = e^ln(a), so a^x = e^(x*ln(a)), and: f'(x) = ln(a)*e^(x*ln(a)) = a^x*ln(a) g'(x) = 1 Putting this together: lim (a^x - 1)/x = lim a^x*ln(a)/1 = ln(a) x->0 x->0 3) lim [ln(1 + x)]/x = lim [1/(x + 1)]/1 = 1 x->0 x->0 If you want a proof of L'Ho^pital's Rule, either wait until you get to that point in calculus, or write again. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 05/27/98 at 15:18:34 From: Kumarpal Sheth Subject: Re: Proofs for limits Hello Dr. Math, Thanks a lot for the proof. I didn't understand everything but I will try and figure it out. I'll let you know if I have further questions. Thanks a lot once again. Kumarpal Sheth |
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