Find $\lim_{n\to \infty}(\cos{x\over n})^{n^2}$ Find $$\lim_{n\to \infty}\left(\cos{x\over n}\right)^{n^2}$$ where $x\in \Bbb{R}.$ I tried using taylor series. A complete mess, and an area I am not very good at. I tried using $e$ which also gave me nothing. I am lost a little. I would appreciate your help, thorough or hinted. 
 A: Taylor expansion should work. Since the argument of the cosine approaches zero in the given limit, we can neglect all higher-order terms and are left with
$\cos \frac{x}{n} \approx 1 - \frac{x^2}{2n^2}$
We can now write
$(cos \frac{x}{n})^{n^2} = \left[ \left(1 - \frac{x^2}{2n^2}\right)^\frac{2n^2}{x^2}\right]^\frac{x^2}{2} \to e^{-\frac{x^2}{2}}$ in the limit of $n\to\infty$
A: Combine your two ideas!
This demonstration is not altogether rigorous but it does get the right answer:
Assume $x$ is not a multiple of $\pi$. Taylor expand in $\frac{x}{n}$:
$$
\cos \frac{x}{n} = \left( 1 - \frac{x^2}{2n^2} + O(n^{-4}) \right)
$$
Then 
$$
\lim_{n\rightarrow\infty} \left( \cos \frac{x}{n} \right)^{n^2} = \lim_{n\rightarrow\infty} \left(1 - \frac{x^2}{2n^2} \right)^{n^2} =
\lim_{n\rightarrow\infty} \left(1 - \frac{x^2}{2m} \right)^{m} = e^{-x^2/2}
$$
However, if $x = 2k\pi$ then the limit is $1$ since for all integer $n$ the value is $1$.  And for $x = (2k+1)\pi$ the limit does not exist.
A: Noting that $\sin^2\left(\frac xn\right)=1-\cos^2\left(\frac xn\right)$ yields
$$
\cos\left(\frac xn\right)=1-\frac{\sin^2\left(\frac xn\right)}{1+\cos\left(\frac xn\right)}
$$
Therefore,
$$
\begin{align}
\left[\cos\left(\frac xn\right)\right]^{n^2}
&=\left[1-\frac{n^2\sin^2\left(\frac xn\right)}{1+\cos\left(\frac xn\right)}\frac1{n^2}\right]^{n^2}\\
&=\left[1-\frac{a_n}{n^2}\right]^{n^2}
\end{align}
$$
where $a_n=\dfrac{n^2\sin^2\left(\frac xn\right)}{1+\cos\left(\frac xn\right)}$ and $\lim\limits_{n\to\infty}a_n=\dfrac{x^2}2$.
Since $\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n=e^x$ and the convergence is uniform on compact sets, we get that
$$
\begin{align}
\lim_{n\to\infty}\left[\cos\left(\frac xn\right)\right]^{n^2}
&=\lim_{n\to\infty}\left[1-\frac{a_n}{n^2}\right]^{n^2}\\
&=\lim_{n\to\infty}\left[1-\frac{x^2}{2n^2}\right]^{n^2}\\[9pt]
&=e^{-x^2/2}
\end{align}
$$
A: If $x=0$ there's nothing to prove: the limit is $1$. Exchanging $x$ with $-x$ gives the same sequence, so we can assume $x>0$.
It's generally better to compute the limit of the logarithm of the sequence:
$$
\lim_{n\to\infty}n^2\log\cos\frac{x}{n}.
$$
If the limit
$$
\lim_{y\to\infty} y^2\log\cos\frac{x}{y}
$$
exists, then also our limit on the natural numbers exists and they're equal. Now we can do the substitution $t=x/y$, so the limit becomes
$$
\lim_{t\to0^+}x^2\frac{\log\cos t}{t^2}.
$$
Leaving aside the constant $x^2$, we can apply l'Hôpital's theorem:
$$
\lim_{t\to0^+}\frac{\log\cos t}{t^2}=
\lim_{t\to0^+}\frac{-\sin t/\cos t}{2t}=
\lim_{t\to0^+}-\frac{\sin t}{t}\frac{1}{2\cos t}=-\frac{1}{2}.
$$
Thus
$$
\lim_{n\to\infty}n^2\log\cos\frac{x}{n}=-\frac{x^2}{2}
$$
and the original limit is $e^{-x^2/2}$ (which also holds for $x\le0$ as we saw at the beginning).
Of course one can also do
$$
\lim_{t\to0^+}\frac{\log\cos t}{t^2}=
\lim_{t\to0^+}\frac{\log(1-t^2/2+o(t^2))}{t^2}=
\lim_{t\to0^+}\frac{-t^2/2+o(t^2)}{t^2}=-\frac{1}{2}.
$$
