# Limits question L'Hopitals rule

1. Nov 15, 2007

### Sags

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

2. Nov 15, 2007

### Dick

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.

3. Nov 15, 2007

### Sags

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)

Last edited: Nov 15, 2007
4. Nov 15, 2007

### HallsofIvy

Staff Emeritus
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
$$\frac{xcos(x)- sin(x)}{2x^2sin(x)$$
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)'!