1.
$\displaystyle \lim_{x\to 0}(\frac{\sin{x}}{x})^\frac{1}{x}$
2.
$\displaystyle \lim_{x\to 0}(\frac{\sin{3x}}{3x})^\frac{1}{x}$
Should I use just L'Hopital rules or Taylor's teorem?
Consider this, as $\displaystyle x\to{0}$ we have that $\displaystyle \sin(x)\to{x}$
So let us rewrite this limit as follows
$\displaystyle \begin{aligned}\lim_{x\to{0}}\frac{x\cos(x)-x}{x\sin(x)}&=\lim_{x\to{0}}\frac{x\cos(x)-x}{x^2}\\
&=\lim_{x\to{0}}\frac{1-\cos(x)}{x}
\end{aligned}$
Now this is a simple application of L'hopital's,