Limit exercise mixed with integral and trignonemetry asked by Dan Sitaru in the calculus4u group

Exercise:

Find $\underset{n\to \infty }{\mathop{\lim }}\,{{n}^{2}}\int_{0}^{1}{\frac{\sin x}{{{e}^{nx}}}dx}$

Solution: Let ${{I}_{n}}=\int_{0}^{1}{\frac{\sin x}{{{e}^{nx}}}dx}=\int_{0}^{1}{{{e}^{-nx}}\sin x\,dx}$

Let $u={{e}^{-nx}}\,\,\And \,\,dv=\sin x\,dx\Leftrightarrow du=-n{{e}^{-nx}}dx\,\,\And \,\,v=-\cos x$

${{I}_{n}}=\int_{0}^{1}{{{e}^{-nx}}\sin x}\,dx=\left( -{{e}^{-nx}}\cos x \right)_{0}^{1}-n\int_{0}^{1}{{{e}^{-nx}}\cos x}\,dx$

Let $u={{e}^{-nx}}\,\,\And \,\,dv=\cos x\,dx\Leftrightarrow du=-n{{e}^{-nx}}\,\,\And \,v=\sin x$

So \(\int_{0}^{1}{{{e}^{-nx}}\cos x\,dx}=\left( {{e}^{-nx}}\sin x \right)_{0}^{1}+n\int_{0}^{1}{{{e}^{-nx}}\sin x}\,dx\)

Hence, ${{I}_{n}}=\left( -{{e}^{-nx}}\cos x \right)_{0}^{1}-n\left( {{e}^{-nx}}\sin x \right)_{0}^{1}-{{n}^{2}}\int_{0}^{1}{{{e}^{-nx}}\sin xdx}$

$\Leftrightarrow \left( 1+{{n}^{2}} \right){{I}_{n}}=\left( -{{e}^{-nx}}\left( \cos x+n\sin x \right) \right)_{0}^{1}$

$\Leftrightarrow \int_{0}^{1}{\frac{\sin x}{{{e}^{nx}}}dx}=-\left[ \frac{{{e}^{-nx}}\left( \cos x+n\sin x \right)}{{{n}^{2}}+1} \right]_{0}^{1}$

$\Leftrightarrow \int_{0}^{1}{\frac{{{n}^{2}}\sin x}{{{e}^{nx}}}dx}=-\frac{{{n}^{2}}}{{{n}^{2}}+1}\left( \frac{\cos x+n\sin x}{{{e}^{nx}}} \right)_{0}^{1}$

$\Leftrightarrow \int_{0}^{1}{\frac{{{n}^{2}}\sin x}{{{e}^{nx}}}dx}=-\frac{{{n}^{2}}}{{{n}^{2}}+1}\left( \frac{\cos 1+\sin 1}{{{e}^{n}}}-\frac{\cos 0}{{{e}^{0}}} \right)=\frac{{{n}^{2}}}{{{n}^{2}}+1}\left( 1+\frac{\cos 1+\sin 1}{{{e}^{n}}} \right)$

as $n\to \infty$ , $\int_{0}^{1}{\frac{{{n}^{2}}\sin x}{{{e}^{nx}}}dx}\to 1$  ($\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{n}^{2}}}{{{n}^{2}}+1}=1\,\,\And \,\underset{n\to \infty }{\mathop{\lim }}\,\frac{\cos 1+\sin 1}{{{e}^{n}}}=\frac{1}{\infty }=0$ )