TITLE: Can you show me a good approach for taking the limit of this function?
QUESTION [6 upvotes]: I tried to use binomial expansion, but I didn't get the same result.
I would like to know how to approach this please.
I know the answer is $\sqrt{e}$.
My problem is :
$$\lim\limits_{x\to 0} \left(1+\frac{1-\cos x}{x}\right)^{\frac{1}{x}}$$
and also, can I find some good manipulation with this kind of problems ?

REPLY [8 votes]: You can use this way:
\begin{eqnarray}
\lim\limits_{x\to 0} \left(1+\frac{1-\cos x}{x}\right)^{\frac{1}{x}}&=&\lim\limits_{x\to 0} \left(\left(1+\frac{1-\cos x}{x}\right)^{\frac{x}{1-\cos x}}\right)^{\frac{1-\cos x}{x^2}}=e^{\frac12}
\end{eqnarray}
This is because
$$ \lim_{x\to0}\frac{1-\cos x}{x}=0,  \lim_{x\to0}\frac{x}{1-\cos x}=\infty, \lim_{x\to0}(1+x)^{\frac1x}=e,\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12. $$