Finding the limit : $\lim_{x\to0}\ln(e + 2x)^\frac{1}{\sin x}$ I tried replacing $x$ with $0$ the log returns $1$ and the $1/\sin x$ returns $1/0$. So I thought the limit should be infinity. However, graphing the function yields undefined value at $0$, and the result shows that the limit is  $e^{2\,e^ {- 1 }}$. I have no idea how did they reach that conclusion.
 A: 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 $e^{2/e}$.
A: $$\lim_{x\to0}\left(\ln(e+2x)\right)^{1/\sin x}$$
$$=\lim_{x\to0}\left(1+\ln\left(1+\dfrac{2x}e\right)\right)^{1/\sin x}$$
$$=\left(\lim_{x\to0}\left(1+\ln\left(1+\dfrac{2x}e\right)\right)^{\dfrac1{\ln\left(1+\dfrac{2x}e\right)}}\right)^{\lim_{x\to0}\dfrac{\ln\left(1+\dfrac{2x}e\right)}{\sin x}}$$
The inner limit converges to $e$ as $\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$
Now for the exponent
$$\lim_{x\to0}\dfrac{\ln\left(1+\dfrac{2x}e\right)}{\sin x}=\dfrac 2e\cdot\lim_{x\to0}\dfrac{\ln\left(1+\dfrac{2x}e\right)}{\dfrac{2x}e}\cdot\lim_{x\to0}\dfrac x{\sin x}=?$$
A: We have $ \ln (e+2x)^{1/ \sin x}= e^{\frac{\ln (e+2x)}{\sin x}}.$
Now compute $ \lim_{x \to 0}\frac{\ln (e+2x)}{\sin x}$ with l'Hospital.
