Finding $\lim\limits_{x\to 0} (\sqrt{4+x}-1)^{1/(e^x-1)}$ I have to find the limit of $(\sqrt{4+x}-1)^{1/(e^x-1)}$ as $x\to0$ without de l'Hopital's rule and only with notable limits

For $x\to 0, \operatorname{e}^x\sim 1+x$ and $\sqrt{4+x}=2\left(1+\frac{x}{4}\right)^{\frac{1}{2}}\sim 2\left(1+\frac{x}{8}\right)$ so that
$$
\left(\sqrt{4+x}-1\right)^{1}{\operatorname{e}^x-1}\sim \left(1+\frac{x}{4}\right)^{\frac{1}{x}}
$$
Putting $y=\frac{1}{x}$ the limit is
$$
\lim_{y\to\infty}\left(1+\frac{1/4}{y}\right)^y=\operatorname{e}^{1/4}
$$
