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
 A: 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
$$
\lim_{y\to\infty}\left(1+\frac{1/4}{y}\right)^y=\operatorname{e}^{1/4}
$$
A: Taking the logarithm you can compute
$$
\lim_{x\to 0}\frac{\log(\sqrt{4+x}-1)}{e^x-1}=
\lim_{x\to 0}\frac{\log(\sqrt{4+x}-1)}{x}\frac{x}{e^x-1}
$$
The limit of the second fraction is $1$, so you're reduced to computing
$$
\lim_{x\to 0}\frac{\log(\sqrt{4+x}-1)}{x}
$$
Set $1+u=\sqrt{4+x}-1$, so $\sqrt{4+x}=u+2$ and $4+x=u^2+4u+4$, hence $x=u^2+4u$; when $x\to0$ also $u\to0$, so the limit becomes
$$
\lim_{u\to 0}\frac{\log(1+u)}{u(u+4)}
$$
Can you go from here?

Note: in the computation the two fundamental limits
\begin{gather}
\lim_{x\to 0}\frac{e^x-1}{x}=1\\[1ex]
\lim_{x\to 0}\frac{\log(1+x)}{x}=1
\end{gather}
are used. Check if you are allowed to use them.
A: I'm not totally sure where this is heading, but here's a possible first step.
$$
L = \lim_{x\to0} (\sqrt{4+x}-1)^{\large \frac{1}{e^x-1}}
$$
Multiplying both sides, we have
$$
\begin{align}
\left[\lim_{x \to 0} (\sqrt{4+x}+1)^{\large \frac{1}{e^x-1}}\right] L &= 
\lim_{x\to0} [(\sqrt{4+x}-1)(\sqrt{4+x}+1)]^{\large \frac{1}{e^x-1}}\\
\left[\lim_{x \to 0} (\sqrt{4+x}+1)^{\large \frac{1}{e^x-1}}\right] \cdot L&= 
\lim_{x\to 0}[3+x]^{\large \frac{1}{e^x-1}}
\end{align}
$$
Hopefully that helps.
A: $\ln \left( \sqrt{4+x}-1 \right)^{\frac{1}{e^x-1}}= \frac{x}{e^x-1} \frac{\ln \left(1+(\sqrt{4+x}-2)) \right)}{x} = \frac{x}{e^x-1} \frac{\ln \left(1+ \frac{x}{2+\sqrt{4+x}} \right)}{x} = \frac{x}{e^x-1} \frac{\ln \left(1+ \frac{x}{2+\sqrt{4+x}} \right)}{\frac{x}{2+\sqrt{4+x}}} \frac{1}{2+\sqrt{4+x}} \rightarrow 1 \cdot 1 \cdot \frac{1}{4}$
Therefore $\lim_{x\to 0} \left( \sqrt{4+x}-1\right)^\frac{1}{e^x-1} = e^\frac{1}{4}$
I used well known limits $\lim_{x\to 0} \frac{e^x-1}{x}=1=\lim_{x\to 0} \frac{\ln (1+x)}{x}$
