Prove that $\lim \limits_{n \to \infty} n^2 \int_{0}^{\frac{1}{n}} x^{x+1} dx = \dfrac{1}{2} $ Prove that 
$$\lim \limits_{n \to \infty} n^2 \int_{0}^{\frac{1}{n}} x^{x+1} dx = \dfrac{1}{2} $$
 A: My favorite technique, start with the Taylor expansion of $x^{x+1}$ around $x=0$. I am only carrying the first two terms because doing this by hand can get very tedious. And the steps can all be rigorously justified.
\begin{eqnarray*}
x^{x+1}&=&x+\ln(x) x^2+\cdots \\
\int_0^{1/n} x^{x+1} dx&=&\frac{1}{2n^2}-\frac{1+3\ln(n)}{9n^3}+\cdots\\
n^2 \int_0^{1/n} x^{x+1}dx&=&\frac{1}{2}-\frac{1+3\ln(n)}{9n}+\cdots\\
\lim_{n\rightarrow \infty}n^2 \int_0^{1/n} x^{x+1}dx&=&\frac{1}{2}.
\end{eqnarray*}
A: Scale the integral:  substitute $u = n x$ and get
$$\begin{align}n^2 \int_0^{1/n} dx \: x^{x+1} &= \int_0^1 du \: u \, \left (\frac{u}{n}\right )^{u/n}\\ &= \int_0^1 du \: u \, e^{(u/n) \log{(u/n)}}\\ \end{align}$$
As $n \rightarrow \infty$, the exponential $\rightarrow 1$ because $u$ is bounded.  Therefore the limit is
$$\int_0^1 du \, u = \frac{1}{2}$$
Taking the limit into the integral is justified so long as the sequence
$$e^{(u/n) \log{(u/n)}}$$
converges uniformly to $1$ over $u \in [0,1]$; in other words, so long as $(u/n) \log{(u/n)}$ converges to $0$ uniformly in this interval.  That is, the supremum of $(u/n) \log{(u/n)}$ over  $u \in [0,1]$ converges to $0$ as $n \rightarrow \infty$.  This is true because of the fact that $u$ is in a bounded interval.  
A: Try change of variable $t = x^2$. Then we have
$\displaystyle\lim_{n \to \infty} n^2 \displaystyle\int_0^{\frac{1}{n}} x^{x+1} dx$ = $\frac{1}{2} \displaystyle\lim_{n \to \infty} \frac{\displaystyle\int_0^{\frac{1}{n^2}} \sqrt{y}^{\sqrt{y}} dy}{\frac{1}{n^2}}$
Note the quotient is of the form $\displaystyle\lim_{b \to a} \displaystyle\frac{\int_a^b f(x) dx}{b - a}$ and I suspect that it is $f(a) = 1$.
