TITLE: Prove that $\lim \limits_{n \to \infty} n^2 \int_{0}^{\frac{1}{n}} x^{x+1} dx = \dfrac{1}{2} $
QUESTION [8 upvotes]: Prove that 
$$\lim \limits_{n \to \infty} n^2 \int_{0}^{\frac{1}{n}} x^{x+1} dx = \dfrac{1}{2} $$

REPLY [4 votes]: 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*}