Find $\lim_{n\to\infty} \log f_n(x) = \lim_{n\to\infty} \frac { \log\left(\cos\left(\frac{x}{\sqrt{n}}\right)\right)} {\frac 1 n }$ When applying L'Hopital's rule first I get this:
$\lim_{n\to\infty}\log f_n(x)=\lim_{n\to\infty}-\frac {\sin\big (\frac x
{\sqrt n} \big ) x\sqrt n}{2\cos \big  ( \frac x {\sqrt n}\big )} $
by applying it second time I get; 
$\lim_{n\to\infty} \log f_n(x)=\lim_{n\to\infty}\big (\frac {x\cos \big ( \frac x {\sqrt n} \big ) \sqrt n } {2\sin \big ( \frac x {\sqrt n} \big ) }-\frac n 2 \big )$
So now from here how to get:
$\lim_{n\to\infty}f_n(x)=e ^{-\frac {x^2} 2}$
 A: Alternative method. Without applying L'Hopital, but instead Taylor expansions (around $0$) of standard functions:
$$
\begin{align}
\cos u &= 1-\frac{u^2}{2} + o(u^2)\\
\ln(1+u) &= u +o(u)\end{align}
$$
we get, as $\lim_{n\to\infty}\frac{x}{\sqrt{n}} = 0$ (for any fixed $x$):
$$
\ln\cos\frac{x}{\sqrt{n}} = \ln\left(1-\frac{x^2}{2n} + o\left(\frac{1}{n}\right)\right)\tag{1}$$
and from there, since $\lim_{n\to\infty}\frac{x^2}{2n} = 0$
$$
\ln\cos\frac{x}{\sqrt{n}} = -\frac{x^2}{2n} + o\left(\frac{1}{n}\right)\,.\tag{2}
$$
Therefore, for any fixed $x\in\mathbb{R}$,
$$
\ln f_n(x)=\frac{\ln\cos\frac{x}{\sqrt{n}}}{\frac{1}{n}} = \frac{-\frac{x^2}{2n} + o\left(\frac{1}{n}\right)}{\frac{1}{n}}
= -\frac{x^2}{2} + o\left(1\right) \xrightarrow[n\to\infty]{} \boxed{-\frac{x^2}{2}}.
$$
By continuity of $\exp$, this immediately implies that
$$
\boxed{\lim_{n\to\infty}f_n(x) = e^{-\frac{x^2}{2}}}
$$
A: We have $f_n(x)={\left(\cos{\left(\frac{x}{\sqrt{n}}\right)}\right)}^n$. Thus, $\ln{f_n(x)}=n\ln{\left(\cos{\left(\frac{x}{\sqrt{n}}\right)}\right)}$ and:
$$\ln{L(x)}:=\ln{\left(\lim_{n\to\infty}f_n(x)\right)}=\lim_{n\to\infty}{n\ln{\left(\cos{\left(\frac{x}{\sqrt{n}}\right)}\right)}}$$
Using the substitution suggested by daulomb, we have $t=\frac{1}{\sqrt{n}}$, so as $n$ approaches $\infty$, $t$ approaches $0$. Thus:
$$\ln{L(x)}=\lim_{t\to0}{\frac{1}{t^2}\ln{\left(\cos{\left(tx\right)}\right)}}$$
Now we use L'Hopital's rule, since we have a $0/0$ indeterminate form (remember that we take the derivative with respect to $t$, not $x$):
$$\ln{L(x)}=\lim_{t\to0}{\frac{\frac{-x\sin{\left(tx\right)}}{\cos{\left(tx\right)}}}{2t}}=-x\lim_{t\to0}{\frac{\tan{\left(tx\right)}}{2t}}$$
We took the $-x$ outside of the limit because it doesn't depend on $t$. Now we apply L'Hopital's rule again:
$$\ln{L(x)}=-x\lim_{t\to0}{\frac{x\sec^2{\left(tx\right)}}{2}}=-\frac{x^2}{2}\lim_{t\to0}{\sec^2{\left(tx\right)}}=-\frac{x^2}{2}$$
thus $L(x)=e^{-\frac{x^2}{2}}$.
