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!
 A: Use $$\log\left[\cos\left(\frac{1}{n}\right)-\sin\left(\frac{1}{n}\right)\right]^n = n\log\left[\cos\left(\frac{1}{n}\right)-\sin\left(\frac{1}{n}\right)\right] = \frac{\log\left[\cos\left(\frac{1}{n}\right)-\sin\left(\frac{1}{n}\right)\right]}{\frac{1}{n}}$$
and then apply l'Hospital's rule.
A: You may consider 


*

*$\left(\frac{\cos x - \sin x}{1-x}\right)^{\frac{1}{x}}$ for $x\to 0$ and use

*$\lim_{t\to 0} (1+t)^{\frac{1}{t}}= e$
\begin{eqnarray*} \left(\frac{\cos x - \sin x}{1-x}\right)^{\frac{1}{x}}
& = & \left( \underbrace{\left(1 + \frac{\cos x - \sin x -1 +x}{1-x}\right)^{\frac{1-x}{\cos x - \sin x -1 +x}}}_{\stackrel{x\to 0}{\longrightarrow}e}\right)^{\underbrace{\frac{\cos x - \sin x -1 +x}{x-x^2}}_{\stackrel{L'Hop}{\sim}\frac{-\sin x - \cos x +1}{1-2x}\stackrel{x\to 0}{\longrightarrow}0}}\\
& \stackrel{x\to 0}{\longrightarrow} & e^0 = 1
\end{eqnarray*}
A: You can get rid of the trigonometric functions by rewriting the exponent $n$ using
$$\frac1n=\dfrac{\dfrac1n}{\sin\dfrac1n}\sin\dfrac1n.$$
As the larger fraction tends to $1$, it can be ignored.
Now, with $t:=\sin\dfrac1n$, we have
$$\lim_{t\to0}\left(\sqrt{1-t^2}-t\right)^{1/t}.$$
Using Taylor, 
$$\sqrt{1-t^2}-t=1-t+o(t)$$ and the limit is the same as
$$\lim_{t\to0}\left(1-t\right)^{1/t}.$$

Or by L'Hospital on the logarithm,
$$\frac{\log(\sqrt{1-t^2}-t)}t\to\frac{-\dfrac t{\sqrt{1-t^2}}-1}{\sqrt{1-t^2}-t}\to-1.$$
A: Since we recall that
$$\sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots $$
and
$$\cos x=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots$$
for all $x$, then
$$\cos\left(\frac1n\right)-\sin\left(\frac1n\right)=1-\frac1n+\dots$$
and so
$$\lim\limits_{n\to \infty} \left(\cos\left(\frac1n\right)-\sin\left(\frac1n\right) \right) ^n=\lim\limits_{n\to \infty} \left(1- \frac1n  \right) ^n=e^{-1}\, .$$
A: 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 $\frac{1}{e}$.
A: Using the well known limit $$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n=e\tag{1}$$ it is easy to prove that $$\lim_{n\to\infty} \left(1-\frac{1}{n}\right)^n=\lim_{n\to\infty}\dfrac{1}{\left(1+\dfrac{1}{n-1}\right)^{n-1}}\cdot\frac{n-1}{n}=\frac{1}{e}\tag{2}$$ Next we use the following lemma :

Lemma: If $\{a_n\} $ is sequence such that $n(a_n-1)\to 0$ then $a_n^n\to 1$.

We can now choose $$a_n=\dfrac{\cos\left(\dfrac{1}{n}\right)-\sin\left(\dfrac{1}{n}\right)} {1-\dfrac{1}{n}} $$ and note that $$n(a_n-1)=n\left(\dfrac{n\cos\left(\dfrac{1}{n}\right)-n\sin\left(\dfrac{1}{n}\right)-n+1} {n-1}\right) $$ and the above clearly tends to $0$ so that by lemma above $a_n^n\to 1$ and therefore using $(2)$ the desired limit is $1/e$. 
