Prove that $\lim_{n\to\infty} n^2 \int _0^{1/n}{x^{x+1}}dx = {1\over2}$. Prove that
$$\lim_{n\to\infty} n^2 \int _0^{1/n}{x^{x+1}}dx = {1\over2}.$$
My Attempt:
When I saw the integral it was tempting to see the integrand as  $x^x x$ and to think of $lim_{x\to0}$ $x^x$ which when evaluated using L-Hopital's gives 1, will it be helpful in reaching a desired conclusion?

$\newcommand{\d}{\,\mathrm{d}}$If you enforce $x=t/n$, you get: $$\lim_{n\to\infty}n^2\int_0^{1/n}x^{x+1}\d x=\lim_{n\to\infty}\int_0^1t^{1+t/n}n^{-t/n}\d t$$We have: $$0\le t^{1+t/n}n^{-t/n}\le1$$The dominated convergence theorem is then applicable: It is also the case that $1\ge n^{-t/n}\ge n^{-1/n}\to1$ as $n\to\infty$, so in pointwise evaluation of the integrand (justified by DCT) we find: $$\lim_{n\to\infty}\int_0^1t^{1+t/n}n^{t/n}\d t=\int_0^1 t^1\d t=\frac{1}{2}$$
EDIT: A much cleaner way to prove this directly without appellation to the DCT is as follows: $$t\ge t^{1+t/n}n^{-t/n}\ge t^{1+1/n}n^{-1/n}$$On the desired interval, so we have: $$\begin{align}\frac{1}{2}&=n^{1/n}\int_0^1t\d t\ge\int_0^1t^{1+t/n}n^{-t/n}\d t\ge n^{-1/n}\int_0^1t^{1+1/n}\d t\\&= \frac{1}{2+1/n}\cdot n^{-1/n}\end{align}$$From which the squeeze theorem makes an obvious limit of $1/2$.
