How to find $\lim\limits_{n \to \infty}n\int_{0}^{1}(\cos x-\sin x)^ndx$? I want to compute 
$$\lim\limits_{n \to \infty}n\int_{0}^{1}(\cos x-\sin x)^ndx$$
Someone helped me find the limit of the integral, which is $0$, but now I can't figure out this one. Also tried squeeze theorem but I only get only one side of it to converge to $1$. Moreover, I'm bound to using elementary calculus. 

Alternative solution
For $n\ge 4$, we have
\begin{align*}
I_n &= n\int_0^{1/\sqrt{n}} (\cos x - \sin x)^n \mathrm{d} x
+ n\int_{1/\sqrt{n}}^{\pi/4} (\cos x - \sin x)^n \mathrm{d} x\\
&\quad + n\int_{\pi/4}^1 (\cos x - \sin x)^n \mathrm{d} x\\
&= I_{n, 1} + I_{n, 2} + I_{n, 3}.
\end{align*}
Let us prove that $\lim_{n\to \infty} I_{n, 1} = 1$, $\lim_{n\to \infty} I_{n, 2} = 0$,
and $\lim_{n\to \infty} I_{n, 3} = 0$.

*

*First, since $|\cos x - \sin x| = \sin x - \cos x \le \sin x \le \sin 1$ for all $x$ in $[\pi/4, 1]$, we have
$$|I_{n, 3}| \le n\int_{\pi/4}^1 |\cos x - \sin x|^n \mathrm{d} x
\le n (\sin 1)^n (1 - \pi/4).$$
Since $\lim_{n\to \infty} n (\sin 1)^n (1 - \pi/4) = 0$, by the squeeze theorem, we have
$$\lim_{n\to \infty} I_{n, 3} = 0.$$


*Second, it is easy to prove that $0\le \cos x - \sin x
\le \cos \frac{1}{\sqrt{n}} - \sin \frac{1}{\sqrt{n}} \le 1 - \frac{1}{\sqrt{n}}$ for
all $n\ge 4$ and all $x$ in $[1/\sqrt{n}, \pi/4]$. Thus, we have
$$0 \le I_{n, 2} \le n\left(1 - \tfrac{1}{\sqrt{n}}\right)^n\frac{\pi}{4}.$$
Since $\lim_{n\to \infty} n\left(1 - \tfrac{1}{\sqrt{n}}\right)^n = 0$, by the squeeze theorem, we have
$$\lim_{n\to \infty} I_{n, 2} = 0.$$


*Third, with the substitution $x = \frac{y}{\sqrt{n}}$, using $(\cos x - \sin x)^2 = 1 - \sin 2x$, we have
$$I_{n, 1}
= \int_0^1 \sqrt{n}\left(1 - \sin \frac{2y}{\sqrt{n}}\right)^{n/2} \mathrm{d} y.$$
It is easy to prove that, for all $y$ in $[0, 1]$ and all $n\ge 4$,
$$\left(1 - \frac{1}{n}\right)\frac{2y}{\sqrt{n}} \le \sin \frac{2y}{\sqrt{n}} \le \frac{2y}{\sqrt{n}}.$$
Thus, we have
$$\int_0^1 \sqrt{n}\left(1 - \frac{2y}{\sqrt{n}}\right)^{n/2} \mathrm{d} y \le I_{n, 1} \le \int_0^1 \sqrt{n}\left(1 -  \left(1 - \frac{1}{n}\right)\frac{2y}{\sqrt{n}}\right)^{n/2} \mathrm{d} y$$
which results in
$$\frac{n \left(1 - \left(1 - \frac{2}{\sqrt{n}}\right)^{n/2 + 1}\right)}{n+2}
\le I_{n, 1} \le \frac{n^2 \left(1 - \left(1 - \left(1 - \frac{1}{n}\right)\frac{2}{\sqrt{n}}\right)^{n/2 + 1}\right)}{(n+2)(n-1)}.$$
Since
\begin{align*}
\lim_{n\to \infty} \frac{n \left(1 - \left(1 - \frac{2}{\sqrt{n}}\right)^{n/2 + 1}\right)}{n+2} &= 1, \\
\lim_{n\to \infty} \frac{n^2 \left(1 - \left(1 - \left(1 - \frac{1}{n}\right)\frac{2}{\sqrt{n}}\right)^{n/2 + 1}\right)}{(n+2)(n-1)} &= 1,
\end{align*}
by the squeeze theorem, we have $\lim_{n\to \infty} I_{n, 1} = 1$.
Thus, we have $\lim_{n\to \infty} I_n = 1$.
We are done.
