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Limits question L'Hopitals rule 
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#1
Nov1507, 08:58 AM

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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
Nov1507, 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.



#3
Nov1507, 09:20 AM

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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) 


#4
Nov1507, 05:40 PM

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Limits question L'Hopitals rule
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] 


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