Evaluate $\displaystyle \lim_{n \to \infty} n \int_0^1 (\cos x - \sin x)^n dx$ Evaluate $$\lim_{n \to \infty} n \int_0^1 (\cos x - \sin x)^n dx$$
The answer should be $1$.
I tried to solve it similar to how one user solved this integral Limit of integral with cos and sin
but it seems like it doesn't work because the upper bound will become $1 - \pi/4$.

If you know how to deal with
$$n \int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx,$$
just split the integral at $\frac{\pi}{4}$ and see whether you can find something useful for the other part.
Since $x \mapsto \cos x - \sin x$ is strictly decreasing on $\bigl[0, \frac{\pi}{2}\bigr]$, and $\cos \frac{\pi}{4} = \sin \frac{\pi}{4}$, we have
$$\lvert \cos x - \sin x\rvert \leqslant c := \sin 1 - \cos 1$$
for $\frac{\pi}{4} \leqslant x \leqslant 1$. Hence
$$\Biggl\lvert \int_{\frac{\pi}{4}}^1 (\cos x - \sin x)^n\,dx\Biggr\rvert \leqslant \int_{\frac{\pi}{4}}^1 c^n\,dx < c^n.$$
Since $0 < c < 1$, it follows that
$$n \int_{\frac{\pi}{4}}^1 (\cos x - \sin x)^n\,dx \to 0.$$
That is useful indeed. Thus we need only consider
$$\int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx = 2^{n/2} \int_0^{\frac{\pi}{4}} \sin^n \biggl(\frac{\pi}{4} - x\biggr)\,dx = 2^{n/2}\int_0^{\frac{\pi}{4}} \sin^n x\,dx.$$
Substituting $u = \sin x$, then $v = \sqrt{2}\cdot u$, and afterwards integrating by parts, we find (for $n \geqslant 1$)
\begin{align}
n\int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx
&= n 2^{n/2} \int_0^{\frac{\pi}{4}} \sin^n x\,dx \\
&= n 2^{n/2} \int_0^{\frac{1}{\sqrt{2}}} \frac{u^n}{\sqrt{1-u^2}}\,du \\
&= n \int_0^1 \frac{v^n}{\sqrt{2-v^2}}\,dv \\
&= v^n\cdot \frac{v}{\sqrt{2-v^2}}\biggr\rvert_0^1 - \int_0^1 v^n\biggl(\frac{1}{\sqrt{2-v^2}} + \frac{v^2}{(2-v^2)^{3/2}}\biggr)\,dv \\
&= 1 - \int_0^1 \frac{2v^n}{(2-v^2)^{3/2}}\,dv.
\end{align}
Since $\frac{1}{\sqrt{2}} < \frac{2}{(2-v^2)^{3/2}} < 2$ for $0 < v < 1$, we thus have
$$1 - \frac{2}{n+1} < n\int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx < 1 - \frac{1}{\sqrt{2}\,(n+1)}.$$
