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Limits question L'Hopitals rule

by Sags
Tags: lhopitals, limits, rule
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Sags
#1
Nov15-07, 08:58 AM
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
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Dick
#2
Nov15-07, 09:13 AM
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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.
Sags
#3
Nov15-07, 09:20 AM
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)

HallsofIvy
#4
Nov15-07, 05:40 PM
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Limits question L'Hopitals rule

Quote Quote by Sags View Post
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
[tex]\frac{\frac{cos(x)}{sin(x)}- 1/x}}{2x}[/tex]
so I assume the "1/y" was "1/x". That reduces to
[tex]\frac{xcos(x)- sin(x)}{2x^2sin(x)[/tex]
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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