Evaluating $\lim_{n\rightarrow \infty}n\int_{0}^{1}\left(\cos x-\sin x\right)^n\text{ d}x$ $$\lim_{n\rightarrow \infty}n\int_{0}^{1}\left(\cos x-\sin x\right)^n\text{ d}x$$
A) $ \infty$
B) $ 0$
C) $ 1$
D) $ \frac{1}{2}$
E) $\cos 1$
Source: admission 2015 Technical University of Cluj-Napoca
I tried with this result, but I'm not if it holds
If $f:[a,b]\rightarrow R$ f is differentiable with derivative nonzero, and $\displaystyle \lim_{n\to \infty}\int_a^b f^n(x)dx=0$  then $\displaystyle \lim_{n\to\infty} n \cdot \int_a^b f^n(x)dx=\lim_{n\to\infty} \left(\frac{f^n(b)}{f'(b)}-\frac{f^n(a)}{f'(a)}\right)$
I used $f(x) = cosx - sinx $ but I'm not sure if $\displaystyle \lim_{n\to \infty}\int_a^b f^n(x)dx=0$ 
I have tried with this too, $\cos x-\sin x= \frac{2}{\sqrt{2}}\cos (x+\frac{\pi}{4})$, but at one moment I can't continue... 
I can find a recurrence formula, but it's very complicated and I can't see the limit...
 A: Let $f(x)=\cos x-\sin x$.  We note that for $x\in[0,1]$, the maximum of $f(x)$ occurs at $x=0$ for which $f(0)=1$.  
And for any number $0<a < \pi/4$, $x\in [a,\pi/4]$, $0<f(x)<\cos( a) -\sin (a)<1$.
Now, we choose a positive number $\delta<\pi/2-1$.  Then, we write the integral of interest as the sum
$$\begin{align}
\int_0^1 (\cos x-\sin x)^n\,dx&=\int_{0}^{\delta} (\cos x-\sin x)^n\,dx+\int_{\delta}^{\pi/2-1} (\cos x-\sin x)^n\,dx\\\\
&+\int_{\pi/2-1}^{\pi/4}(\cos x-\sin x)^n\,dx+\int_{\pi/4}^{1}(\cos x-\sin x)^n\,dx\\\\
&=\int_{0}^{\delta} (\cos x-\sin x)^n\,dx+\int_{\delta}^{\pi/2-1} (\cos x-\sin x)^n\,dx \tag 1\\\\
&+(1+(-1)^n)\int_{\pi/2-1}^{\pi/4}(\cos x-\sin x)^n\,dx 
\end{align}$$

Now, we can see that the second and third integrals on the right-hand side of $(1)$ are positive and bounded by $(1-\pi/4)(\cos \delta-\sin \delta)^n$ and $2(1-\pi/4)(\sin (1)-\cos (1))^n$, respectively.  Therefore, 
$$\lim_{n\to \infty}n\int_{\delta}^1(\cos x-\sin x)^n\,dx=0$$

For the first integral on the right-hand side of $(1)$, we proceed as follows.  Given any $\epsilon>0$, we now choose $0<\delta<\epsilon$.  Then, using the mean-value theorem, we find that for $x\in[0,\delta]$, the function $f$ satisfies the inequalities 
$$1-(1+\epsilon)x \le f(x)\le 1-x \tag 2$$  
Applying $(2)$ to the first integral on the right-hand side of $(1)$ yields
$$n\int_{0}^{\delta}\left(1-(1+\epsilon)x\right)^n\,dx\le n\int_{0}^{\delta} (\cos x-\sin x)^n\,dx\le n\int_{0}^{\delta} (1- x)^n\,dx \tag 3$$
Evaluating the integrals in $(3)$ that bound $ n\int_{0}^{\delta} (\cos x-\sin x)^n\,dx$ reveals
$$\frac{n}{n+1}\frac{\left(1-(1-(1+\epsilon)\delta)^{n+1}\right)}{1+\epsilon}\le n\int_{0}^{\delta} (\cos x-\sin x)^n\,dx\le \frac{n}{n+1}\left(1-(1-\delta)^{n+1}\right) \tag 3$$
Using the squeeze theorem ($n\to \infty$) we see that for any given $\epsilon>0$, we can choose $0<\delta <\epsilon$ so that 
$$1-\epsilon\le \frac{1}{1+\epsilon}\le \lim_{n\to \infty}n\int_{0}^{\delta} (\cos x-\sin x)^n\,dx\le 1$$
Therefore, we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}n\int_{0}^{\delta} (\cos x-\sin x)^n\,dx=1}$$
