Wolfram Alpha gives me this solution:
$$\lim\limits_{x\to 0} \left(\frac {1+e^{x}}{2}\right) ^{1/x} = \sqrt{e}$$
But I have no idea how to get to that result.
I tried using L'Hopital but I found it impossible to get rid of the 1/x.
I also tried to use the fact that the result is $e^{1/2}$ and put an $e^{ln()}$ to use the logarithms properties, but I still couldn't simplify/rewrite the 1/x
I thought of using the squeeze theorem but I couldn't find functions that could work with the inequalities.
Best Answer
According to the continuity of the $\exp$ function as well as the l'Hôpital's rule, one gets:
\begin{align*} \lim_{x\to 0}\left(\frac{1 + e^{x}}{2}\right)^{1/x} & = \lim_{x\to 0}\exp\left(\frac{1}{x}\ln\left(\frac{1 + e^{x}}{2}\right)\right) = \exp\left(\lim_{x\to 0}\frac{e^{x}}{1 + e^{x}}\right) = \exp\left(\frac{1}{2}\right) = \sqrt{e} \end{align*}
Hopefully this helps!