# Limits question L'Hopitals rule

by Sags
Tags: lhopitals, limits, rule
 P: 2 1. The problem statement, all variables and given/known data lim (sin(x)/x)^(1/X^2) x->0 2. Relevant equations for the life of me i cannot get the correct solution 3. The attempt at a solution Ive tried taking the log of both sides etc and working from there then applying l'hopitals rule until i get a result but the answer i always get is e^(-1/2) but the answer is e^(-1/6) any help would be much appreciated
 Sci Advisor HW Helper Thanks P: 25,235 e^(-1/6) is correct. And that's the way to do it alright. But there is no way to tell why you get e^(-1/2) unless you show us what you did.
 P: 2 sorry ill give more info lim (sin(x)/x)^(1/X^2) x->0 let y = (sin(x)/x)^(1/X^2) ln y = (ln(sin(x)/x))^(1/X^2) ln y = (ln(sin(x)/x))/(X^2) ln y = (ln sin(x) - ln(x))/(X^2) apply l'hopital's rule = (cos(x)/sin(x) - 1/y)/2X apply l'hotital's rule again = (-sin(x)/cos(x) -1/1)/2 and from there i get = -1/2 then y = e^(-1/2)
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Limits question L'Hopitals rule

 Quote by Sags sorry ill give more info lim (sin(x)/x)^(1/X^2) x->0 let y = (sin(x)/x)^(1/X^2) ln y = (ln(sin(x)/x))^(1/X^2) ln y = (ln(sin(x)/x))/(X^2) ln y = (ln sin(x) - ln(x))/(X^2) apply l'hopital's rule = (cos(x)/sin(x) - 1/y)/2X
How did y get over on the right side?
Applying L'Hopital's rule gives you
$$\frac{\frac{cos(x)}{sin(x)}- 1/x}}{2x}$$
so I assume the "1/y" was "1/x". That reduces to
$$\frac{xcos(x)- sin(x)}{2x^2sin(x)$$
 apply l'hotital's rule again = (-sin(x)/cos(x) -1/1)/2
You are also differentiating incorrectly. The derivative of cos(x)/sin(x) is NOT (cos(x))'/(sin(x))' and the derivative of 1/x is NOT1/(x)'!

 and from there i get = -1/2 then y = e^(-1/2)

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