Evaluate a limit involving definite integral Evaluate the following limit:
$$\lim_{n \to \infty} \left[n - n^2 \int_{0}^{\pi/4}(\cos x - \sin x)^n dx\right]$$
I've tried to rewrite the expression as follows:
$$\lim_{n \to \infty} \left[n - n^2 \sqrt{2}^n \int_{0}^{\pi/4}\sin^n \left( \frac{\pi}{4} - x \right) dx\right]$$
However, this doesn't seem to help too much. Thank you!

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Note that
\begin{align}
&\lim_{n \to \infty}\braces{n - n^{2}\int_{0}^{\pi/4}\bracks{\cos\pars{x} - \sin\pars{x}}^{\,n}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{n - 2^{n/2}\,n^{2}
\int_{0}^{\pi/4}\cos^{n}\pars{x + {\pi \over 4}}\,\dd x}\label{1}\tag{1}
\end{align}

When $\ds{n \to \infty}$, the main contribution to the integral comes from values of $\ds{x \gtrsim 0}$ such that it's is a 'candidate' to be evaluated by means of the Laplace Method. Namely,

\begin{align}
\int_{0}^{\pi/4}\cos^{n}\pars{x + {\pi \over 4}}\,\dd x & =
\int_{0}^{\pi/4}\exp\pars{n\ln\pars{\cos\pars{x + {\pi \over 4}}}}\,\dd x
\\[5mm] & \sim
\int_{0}^{\infty}\exp\pars{n\bracks{-\,{\ln\pars{2} \over 2} - x}}
\pars{1 - nx^{2}}\,\dd x
\\[5mm] & = 
2^{-n/2}\,\pars{{1 \over n} - {2 \over n^{2}}}\quad
\mbox{as}\ n \to \infty\label{2}\tag{2}
\end{align}


With \eqref{1} and \eqref{2}:

\begin{align}
&\lim_{n \to \infty}\braces{n - n^{2}\int_{0}^{\pi/4}\bracks{\cos\pars{x} - \sin\pars{x}}^{\,n}\,\dd x} =
\lim_{n \to \infty}\braces{n - 2^{n/2}\,n^{2}
\bracks{2^{-n/2}\,\pars{{1 \over n} - {2 \over n^{2}}}}}
\\[5mm] = &\
\bbx{\ds{2}}
\end{align}
