<s>[INST] 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
 [/INST] $\ln \left( \sqrt{4+x}-1 \right)^{\frac{1}{e^x-1}}= \frac{x}{e^x-1} \frac{\ln \left(1+(\sqrt{4+x}-2)) \right)}{x} = \frac{x}{e^x-1} \frac{\ln \left(1+ \frac{x}{2+\sqrt{4+x}} \right)}{x} = \frac{x}{e^x-1} \frac{\ln \left(1+ \frac{x}{2+\sqrt{4+x}} \right)}{\frac{x}{2+\sqrt{4+x}}} \frac{1}{2+\sqrt{4+x}} \rightarrow 1 \cdot 1 \cdot \frac{1}{4}$
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 $\lim_{x\to 0} \frac{e^x-1}{x}=1=\lim_{x\to 0} \frac{\ln (1+x)}{x}$
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