Originally Posted by

**Ulysses** Hi there. Im new on this forum. And here is the deal, I'm trying to solve this limit:

$\displaystyle \displaystyle\lim_{x \to{0}+}{(\displaystyle\frac{sin(x)}{x})^{\frac{1} {x^2}}}$

I have made an attempt, using logaritm, that I think its the way of solving it, but it didn't simplfied things too much.

$\displaystyle \displaystyle\lim_{x \to{0}+}{e^{\frac{\ln\frac{sin(x)}{x}}{x^2}}}$

$\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(\frac{sin(x)}{x})} {x^2}}$

And then:

$\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{\ln(sin(x))-\ln(x))}{x^2}}$

How should I proceed?

I've tried using L'Hôpitals rule, but I didn't get too far.

$\displaystyle \displaystyle\lim_{x \to{0}+}{\displaystyle\frac{x\tan(x)-1}{2x}}$

From here it would seem to be that the answer is $\displaystyle e^{-\infty}$ (which I think it would be equal to zero) but I used derive to make the calculus, and the límit should give $\displaystyle e^{-1/6}$

Bye there.