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.
$$\lim_{x \to 0} \; (2^x + \sin (3x)) ^{\cot(3x)}$$

Rewritting the limit using the exponential function $e^x$ using the logarthmic identity $e^{\ln x} = x$.
\begin{align}
\lim_{x \to 0} (2^x + \sin(3x))^{\cot(3x)} &= \lim_{x \to 0} \exp(\cot(3x)\ln(2^x + \sin(3x)))\\
&= \exp( \lim_{x \to 0} \cot(3x)\ln(2^x + \sin(3x)))\\
&= \exp(\lim_{x \to 0} \frac{\ln(2^x + \sin(3x))}{\tan(3x)} )\\
\end{align}
Now apply L'Hopital's rule
\begin{align}
\lim_{x \to 0} \frac{\ln(2^x + \sin(3x))}{\tan(3x)} &= \lim_{x \to 0} \frac{\frac{3\cos(3x) + 2^x \ln(2)}{\sin(3x) + 2^x}}{3\sec^2{3x}}\\
&= \frac{3\cos(0) + 2^0 \ln(2)}{(\sin(0) + 2^0)(3\sec^2{0})} =\frac{3 + \ln(2)}{3}
\end{align}
Plugging this in yields $\lim_{x \to 0} (2^x + \sin(3x))^{\cot(3x)} = e^{\frac{3 + \ln(2)}{3}} = e^1e^\frac{\ln(2)}{3} = 2^\frac{1}{3}e = \sqrt[3]{2}e$
