Calculate ${\lim\limits_ {n \to \infty} {\cos(a/n)^{n^2}}}$ I would like to calculate 
$$
\lim_ {n \to \infty} {\cos \left(\frac {a}{n}\right)^{n^2}}
$$
where $n \in \mathbb N$ and $a \in \mathbb R \setminus \{0\}$
The answer should be $e^{-a^2/2}$, but I'm not sure how to calculate it.
 A: Notice, let $\frac 1n=t\implies t\to 0$ as $n\to \infty$ $$\lim_{n\to \infty}\left(\cos\left(\frac{a}{n}\right)\right)^{n^2}=\lim_{t\to 0}\left(\cos\left(at\right)\right)^{1/t^2}$$
$$=\exp\lim_{t\to 0}\frac{1}{t^2}\ln\left(\cos\left(at\right)\right)$$
$$=\exp\lim_{t\to 0}\frac{\underbrace{\ln\left(\cos\left(at\right)\right)}_{\longrightarrow 0}}{\underbrace{t^2}_{\longrightarrow 0}}$$
Using L'Hospital's rule for $\frac 00$ form 
$$=\exp\lim_{t\to 0}\frac{\frac{d}{dt}\left(\ln\left(\cos\left(at\right)\right)\right)}{\frac{d}{dt}(t^2)}$$
$$=\exp\lim_{t\to 0}\frac{\frac{-a\sin (at)}{\cos(at)}}{2t}$$
$$=\exp\left(\frac{(-a^2)}{2}\lim_{t\to 0}\left(\frac{\sin(at)}{at}\right)\cdot\lim_{t\to 0} \frac{1}{\cos (at)}\right)$$
$$=\exp\left(\frac{(-a^2)}{2}\left(1\right)\cdot(1)\right)$$ $$=\color{red}{e^{-a^2/2}}$$
A: HINT:
$$\left(\cos\dfrac an\right)^{n^2}=\left(1-\sin^2\dfrac an\right)^{\dfrac{n^2}2}$$
$$=\left(\left(1-\sin^2\dfrac an\right)^{-\dfrac1{\sin^2\dfrac an}}\right)^{-\sin^2\dfrac an\cdot \dfrac{n^2}2}$$
As $n\to\infty,\dfrac an\to0\implies\sin^2\dfrac an\to0$  so the inner limit converges to $e$
Set $\dfrac an=h,$
$$\lim_{n\to\infty}\sin^2\dfrac an\cdot \dfrac{n^2}2=\dfrac{a^2}2\lim_{h\to0}\dfrac{\sin^2h}{h^2}=?$$
A: Using Taylor series:


*

*First, rewrite with exponentials:
$$
\left(\cos\frac{a}{n}\right)^{n^2} = e^{n^2\ln \cos\frac{a}{n}}
$$

*Then, expand, composing $\cos u = 1-\frac{u^2}{2} + o(u^2)$ when $u\to0$ and $\ln(1+u)=u+o(u)$: (place your mouse over the shaded area to show the derivation)

$$\left(\cos\frac{a}{n}\right)^{n^2} = e^{n^2\ln \left(1-\frac{a^2}{2n^2} +o(\frac{1}{n^2})\right)}= e^{n^2\left(-\frac{a^2}{2n^2} +o(\frac{1}{n^2})\right)}=e^{-\frac{a^2}{2} +o(1)} \xrightarrow[n\to\infty]{} e^{-\frac{a^2}{2}}$$

where the very last step relies on the fact that by continuity of the exponential, $e^{o(1)}\xrightarrow[n\to\infty]{} e^{0}=1$. 
