L'Hopital's RuleDate: 8/11/96 at 17:55:17 From: Anonymous Subject: Limit of x^(1/x) Hi! I have a problem with a limit that can be solved with L'Hopital's Rule because it is part of the first partial, but for which I don't see how to apply L'Hopital's Rule: Limit x^(1/x) x-> Inf I am thankful for any help. Date: 8/11/96 at 18:9:6 From: Doctor Paul Subject: Re: Limit of x^(1/x) Let's assume it is this instead: Lt 1^(1/x) x-> Inf Note that as x goes to infinity, 1/x goes to zero, right? so 1 to the zero power is just 1. I think it's pretty safe to assume that x^(1/x) also goes to 1 as x goes to infinity. Let's prove it: You have: Limit x^(1/x) x-> Inf Let's say that's equal to sum number 'n'. We want to solve for n... Limit x^(1/x) = n x-> Inf Take the natural log of both sides: Limit (1/x)*ln(x) = ln(n) x-> Inf Now we have to evaluate the left side. It's a limit that evaluates to infinity/0. Now we can use L'Hopital's Rule on the lefthand side. Take the derivitive and re-evaluate the limit: Limit (1/x) / 1 = ln(n) x-> Inf Now we can see that the lefthand side evaluates to zero. 0 = ln(n) Exponentiate both sides: e^0 = n so n = 1 There's proof that the limit evaluates to one. I hope this helps. Regards, -Doctor Paul, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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