Evaluate $\lim\limits_{n\to \infty} \left( \cos(1/n)-\sin(1/n) \right) ^n $? How to calculate 
$\lim\limits_{n\to \infty} \left( \cos(1/n)-\sin(1/n) \right) ^n $?
Since $\lim\limits_{n\to \infty} \frac {\cos(1/n)-\sin(1/n) }{1-1/n} = 1  $, I guess that the limit above is $\frac{1}{e}$, but since the form $(\to 1)^{\to \infty}$  is indeterminate, I don't know how to prove it formally.
Thanks!

We have  :
$\begin{align*}
\lim\limits_{n\to \infty} \left( \cos\left(\frac{1}{n}\right)-\sin\left(\frac{1}{n}\right) \right) ^n&=\lim\limits_{n\to \infty}\left(1+\cos\left(\frac{1}{n}\right)-\sin\left(\frac{1}{n}\right)-1\right)^n\\&=\lim\limits_{n\to \infty}\left[\bigg(1+\cos(\frac{1}{n})-\sin(\frac{1}{n})-1\bigg)^{\frac{1}{\cos(\frac{1}{n})-\sin(\frac{1}{n})=1}}\right]^{n(\cos(\frac{1}{n})-\sin(\frac{1}{n})-1)}\\&=e^{\lim\limits_{n\to \infty}n(\cos(\frac{1}{n})-\sin(\frac{1}{n})-1)}\end{align*}$
But 
$\begin{align*}\lim\limits_{n\to \infty}n\left(\cos\left(\frac{1}{n}\right)-\sin\left(\frac{1}{n}\right)-1\right)&=\lim\limits_{n\to \infty}\frac{\cos(\frac{1}{n})-\sin(\frac{1}{n})-1}{\frac{1}{n}}\\&=\lim\limits_{x\to 0}\frac{\cos x-\sin x-1}{x}\\&=\lim\limits_{x\to 0}(-\sin x-\cos x)\\&=-1,\end{align*}$
so your limit equals $\frac{1}{e}$.
