Find the limit $\lim_{x \to 0} (2^x + \sin (3x)) ^{\cot(3x)}$ Please help, I have already tried every thing I can, but nothing works. I have no I idea what to do.
$$\lim_{x \to 0} \; (2^x + \sin (3x)) ^{\cot(3x)}$$

While evaluating limit of expressions of type $\{f(x)\}^{g(x)}$ it is best to take logarithms. Let $L$ be the desired limit so that
\begin{align}
\log L &= \log\left(\lim_{x \to 0}(2^{x} + \sin 3x)^{\cot 3x}\right)\notag\\
&= \lim_{x \to 0}\log(2^{x} + \sin 3x)^{\cot 3x}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0}\cot 3x\log(2^{x} + \sin 3x)\notag\\
&= \lim_{x \to 0}\frac{\log(2^{x} + \sin 3x)}{\tan 3x}\notag\\
&= \lim_{x \to 0}\frac{\log(2^{x} + \sin 3x)}{3x}\cdot\frac{3x}{\tan 3x}\notag\\
&= \frac{1}{3}\lim_{x \to 0}\frac{\log(1 + 2^{x} - 1 + \sin 3x)}{x}\notag\\
&= \frac{1}{3}\lim_{x \to 0}\frac{\log(1 + 2^{x} - 1 + \sin 3x)}{2^{x} - 1 + \sin 3x}\cdot\frac{2^{x} - 1 + \sin 3x}{x}\notag\\
&= \frac{1}{3}\lim_{x \to 0}1\cdot\left(\frac{2^{x} - 1}{x} + 3\cdot\frac{\sin 3x}{3x}\right)\notag\\
&= \frac{1}{3}(\log 2 + 3) = 1 + \frac{\log 2}{3}\notag
\end{align}
and hence $L = 2^{1/3}e$.
