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?

Bye there.

\(\displaystyle \lim_{x \to 0^{+}}\left[\left(\frac{\sin{x}}{x}\right)^{\frac{1}{x^2}}\right] = \lim_{x \to 0^{+}}e^{\ln{\left[\left(\frac{\sin{x}}{x}\right)^{\frac{1}{x^2}}\right]}}\)

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

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

This \(\displaystyle \to \frac{0}{0}\), so you can use L'Hospital's Rule...

\(\displaystyle e^{\lim_{x \to 0^{+}}\frac{\ln{\left(\frac{\sin{x}}{x}\right)}}{x^2}} = e^{\lim_{x \to 0^{+}}\left(\frac{\frac{\frac{x\cos{x} - \sin{x}}{x^2}}{\frac{\sin{x}}{x}}}{2x}\right)}\)

\(\displaystyle = e^{\lim_{x \to 0^{+}}\left(\frac{\frac{x\cos{x} - \sin{x}}{x\sin{x}}}{2x}\right)}\)

\(\displaystyle = e^{\lim_{x \to 0^{+}}\left(\frac{x\cos{x} - \sin{x}}{2x^2\sin{x}}\right)}\).

This still \(\displaystyle \to \frac{0}{0}\) so use L'Hospital's Rule again...

\(\displaystyle e^{\lim_{x \to 0^{+}}\left(\frac{x\cos{x} - \sin{x}}{2x^2\sin{x}}\right)} = e^{\lim_{x \to 0^{+}}\left(\frac{\cos{x} - x\sin{x} - \cos{x}}{2x^2\cos{x} + 4x\sin{x}}\right)}\)

\(\displaystyle = e^{\lim_{x \to 0^{+}}\left(-\frac{\sin{x}}{4\sin{x} + 2x\cos{x}}\right)}\)

This still \(\displaystyle \to \frac{0}{0}\), so use L'Hospital's Rule again...

\(\displaystyle e^{\lim_{x \to 0^{+}}\left(-\frac{\sin{x}}{4\sin{x} + 2x\cos{x}}\right)} = e^{\lim_{x \to 0^{+}}\left(-\frac{\cos{x}}{4\cos{x} + 2x\sin{x} + 2\cos{x}}\right)}\)

\(\displaystyle = e^{-\frac{1}{4 + 2}}\)

\(\displaystyle = e^{-\frac{1}{6}}\).

PHEW!

Edit: Fixed a small mistake.