Prove that $\lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}x^{x+1}dx=\frac{1}{2}.$ 
Question: Prove that $$\lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}x^{x+1}dx=\frac{1}{2}.$$

Solution: Let $$I_n:=n^2\int_0^{\frac{1}{n}}x^{x+1}dx, \forall \in\mathbb{N}.$$ Substituting $nx=t$ in $I_n$, we have $$I_n=n\int_0^1\left(\frac{t}{n}\right)^{1+\frac{t}{n}}dt.$$
Now for all $0\le t\le 1$ and for all $n\in\mathbb{N}, n+t\le n+1\implies 1+\frac{t}{n}\le1+\frac{1}{n}.$ This implies that for all $0\le t\le 1$ and for all $n\in\mathbb{N}$, we have $$\left(\frac{t}{n}\right)^{1+\frac{t}{n}}\ge \left(\frac{t}{n}\right)^{1+\frac{1}{n}}.$$
Therefore, for all $n\in\mathbb{N},$ $$\int_0^1\left(\frac{t}{n}\right)^{1+\frac{t}{n}}dt\ge \int_0^1\left(\frac{t}{n}\right)^{1+\frac{1}{n}}dt=n^{-\left(1+\frac{1}{n}\right)}\frac{n}{2n+1}.$$ This implies that $$I_n\ge n^{-\frac{1}{n}}\frac{n}{2n+1},\forall n\in\mathbb{N}.$$
Next note that for all $0\le t\le 1$ and for all $n\in\mathbb{N}$, $1+\frac{t}{n}>1$, which implies that $\left(\frac{t}{n}\right)^{1+\frac{t}{n}}<\frac{t}{n}.$ Therefore, $$\int_0^1\left(\frac{t}{n}\right)^{1+\frac{t}{n}}dt<\int_0^1\left(\frac{t}{n}\right)dt=\frac{1}{2n}.$$ This implies that $$I_n<\frac{1}{2},\forall n\in\mathbb{N}.$$ Thus, for all $n\in\mathbb{N}$, we have $$n^{-\frac{1}{n}}\frac{n}{2n+1}\le I_n<\frac{1}{2}.$$ Now since $$\lim_{n\to\infty}n^{-\frac{1}{n}}\frac{n}{2n+1}=\frac{1}{2},$$ therefore by Sandwich theorem we can conclude that $$\lim_{n\to\infty}I_n=\frac{1}{2}.$$
Is this solution correct and rigorous enough and is there any other way to solve the problem?

$$ \lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}x^{x+1}\mathrm{d}x = \lim_{n \to \infty} n^2 I_n =\frac{1}{2}. $$
My approach is quite similar to @Sanket. Sandwiching is the basic idea.
We have that $$0 \leqslant x \leqslant \frac{1}{n} \implies 1 \leqslant x + 1 \leqslant 1 + \frac{1}{n} \implies x \geqslant x^{x + 1} \geqslant x^{1 + \frac{1}{n}}  $$
$$ \implies n^2 \int_0^{\frac{1}{n}} x~\mathrm{d}x \geqslant n^2 I_n \geqslant 
n^2 \int_{0}^{\frac{1}{n}} x^{1 + \frac{1}{n}}~\mathrm{d}x \implies \frac{1}{2} \geqslant n^2 I_n \geqslant \frac{1}{n^{1/n}\left(2 + \frac{1}{n}\right)}   $$
$$ \implies \frac{1}{2} \geqslant \lim_{n \to \infty} n^2 I_n \geqslant \lim_{n \to \infty} \frac{1}{n^{1/n}\left(2 + \frac{1}{n}\right)} $$
We'll basically use the standard limit $n^{1/n} \to 1$ as $n \to \infty$. Thus:
$$ \frac{1}{2} \geqslant \lim_{n \to \infty} n^2 I_n \geqslant \frac{1}{2} $$
Thus by sandwich theorem we have that $\lim\limits_{n \to \infty} n^2 I_n $ exists and equals $\frac{1}{2}$.
