Prove that $ \lim_{n \to \infty}\int_0^{n^2} n\sin (x/n)e^{-x^2} dx = 1/2 $

Question

Prove that $$ \lim_{n \to \infty}\int_0^{n^2} n\sin(x/n)e^{-x^2} dx = 1/2.$$

I've tried all of the techniques learned in the course so far and none of them seem to work, so if anyone could help me with this it would be much appreciated. Thanks!

Answer 1

Hint. You may need to show:

\begin{align} |\sin(x/n) |\leq \frac{|x|} {n} \end{align} So now one has \begin{align} 0\leq n|\sin(x/n) |e^{-x^2}\leq |x|e^{-x^2} \end{align} Your integral is: \begin{align} \int^\infty_0 \mathbf{1}_{(0,n^2)}n\sin(x/n)e^{-x^2}\,dx \end{align} Now apply The Dominated Convergence Theorem to conclude.