Finding the limit : $\lim_{x\to0}\ln(e + 2x)^\frac{1}{\sin x}$

Using Taylor polynomials \begin{align*} \log (e + 2x)^{\frac{1}{{\sin x}}} & = \exp \left( {\frac{{\log \log (e + 2x)}}{{\sin x}}} \right) = \exp \left( {\frac{{\log \left( {1 + \log \left( {1 + \frac{{2x}}{e}} \right)} \right)}}{{\sin x}}} \right) \\ & = \exp \left( {\frac{{\log \left( {1 + \frac{{2x}}{e} + \mathcal{O}(x^2 )} \right)}}{{\sin x}}} \right) = \exp \left( {\frac{{\frac{{2x}}{e} + \mathcal{O}(x^2 )}}{{\sin x}}} \right) \\ & = \exp \left( {\frac{2}{e}\frac{x}{{\sin x}}(1 + \mathcal{O}(x))} \right). \end{align*} Thus, the limit is indeed $e^{2/e}$.

$$\lim_{x\to0}\left(\ln(e+2x)\right)^{1/\sin x}$$

$$=\lim_{x\to0}\left(1+\ln\left(1+\dfrac{2x}e\right)\right)^{1/\sin x}$$

$$=\left(\lim_{x\to0}\left(1+\ln\left(1+\dfrac{2x}e\right)\right)^{\dfrac1{\ln\left(1+\dfrac{2x}e\right)}}\right)^{\lim_{x\to0}\dfrac{\ln\left(1+\dfrac{2x}e\right)}{\sin x}}$$

The inner limit converges to $e$ as $\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$

Now for the exponent

$$\lim_{x\to0}\dfrac{\ln\left(1+\dfrac{2x}e\right)}{\sin x}=\dfrac 2e\cdot\lim_{x\to0}\dfrac{\ln\left(1+\dfrac{2x}e\right)}{\dfrac{2x}e}\cdot\lim_{x\to0}\dfrac x{\sin x}=?$$