Alternate way to solve $\lim\limits_{x \to 0} (\sin x) ^x$? My solution:
$$ \lim_{x \to 0} (\sin x)^x = \lim_{x \to 0} e^{(x)(\ln \sin x)} = \exp \left( \lim_{x \to 0} (x) (\ln \sin x) \right)$$
Now we have $\lim_{x \to 0} (x) (\ln\sin x)$.
Now  we  can say that the limit is $0$ as $\ln$, $\sin x$ decreases more slowly than $x$.
Question: Is there any other method to solve the above limit without using the arguments saying one function decreases slower?

Just a small variant on Kavi Rama Murthy's answer, together with a comment on the one-sidedness of the limit:
Note,
$$(\sin x)^x=\left(\sin x\over x\right)^xx^x$$
We have
$$\lim_{x\to0}\left(\sin x\over x\right)^x=1^0=1$$
and
$$\lim_{x\to0^+}x^x=1$$
(from the easy L'Hopital for $x\ln x={\ln x\over1/x}$). Therefore
$$\lim_{x\to0^+}(\sin x)^x=1$$
Remark: I've specified the limit from the right, $x\to0^+$, since $x^x$ and $(\sin x)^x$ are undefined (as real-valued functions) for (small) negative values of $x$. The function $\left(\sin x\over x\right)^x$ is defined for negative values of $x$ near $0$, so it's OK to consider its limit as $x\to0$ from both sides.
