Can you show me a good approach for taking the limit of this function? 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 ?
 A: 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
$$ \lim_{x\to0}\frac{1-\cos x}{x}=0,  \lim_{x\to0}\frac{x}{1-\cos x}=\infty, \lim_{x\to0}(1+x)^{\frac1x}=e,\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12. $$
A: $$\frac{1-\cos x}x=\frac x2+o(x)$$
hence
$$\ln\Bigl(1+\frac{1-\cos x}x\Bigr)^{\!\tfrac 1x}=\frac1x\ln\Bigl(1+\frac x2+o(x)\Bigr)=\frac1x\Bigl(\frac x2+o(x)\Bigr)=\frac12+o(1)$$
so that the limit is $\,\mathrm e^{1/2}$.
A: Note that we can write
\begin{equation*}
\lim_{x\to 0}\exp(\frac{\ln(1+\frac{1-\cos(x)}{x})}{x}).
\end{equation*}
Applying L'Hopital's rule twice gives us 
\begin{equation*}
\exp(\lim_{x\to 0}\frac{x\cos(x)}{1+2x-\cos(x)+x\sin(x)}) \\
=\exp(\lim_{x\to 0}\frac{x}{1+2x-\cos(x)+x\sin(x)}).
\end{equation*}
Applying L'Hopital's rule again gives
\begin{equation*}
\exp(\lim_{x\to 0} \frac{1}{2+x\cos(x)+2\sin(x)})=\sqrt{e}~_{\square} 
\end{equation*}
A: Hint look at the $\log$ then use $\cos{x}=1-\frac{x^2}{2}+o(x^3)$ and $\log(1+x)=x-\frac{x^2}{2}+o(x^2)$
Let's look in more details
$$\log\left(\left(1+\frac{1-\cos x}{x}\right)^{\frac{1}{x}}\right)=\frac{1}{x}\log\left(1+\frac{1-\cos x}{x}\right)$$
Using the expansion of $\cos$ we get
$$\log\left(\left(1+\frac{1-\cos x}{x}\right)^{\frac{1}{x}}\right)=\frac{1}{x}\log\left(1+\frac{x}{2}+o(x)\right)$$
And using the expansion of $\log$ we get
$$\log\left(\left(1+\frac{1-\cos x}{x}\right)^{\frac{1}{x}}\right)=\frac{1}{2}+o(x)$$
