Prove that $\lim\limits_{n\to\infty} n^2 \int_0^{1/n} x^{x+1} dx = 1/2.$ 
Prove that $\lim\limits_{n\to\infty} n^2 \int_0^{1/n} x^{x+1} dx = 1/2.$

Suppose we've shown that $\lim\limits_{n\to\infty} n^2\int_0^{1/n} x(x^x-1) dx = 0.$ Then the desired limit equals $\lim\limits_{n\to\infty} n^2 \int_0^{1/n} xdx + n^2\int_0^{1/n} x(x^x-1) dx = \lim\limits_{n\to\infty} n^2 \int_0^{1/n} xdx = 1/2.$
Let $0<a\leq 1$. First, consider evaluating the integral $I_{m,n}(a) := \int_0^a x^m (\ln x)^n dx$ for some $m,n\ge 0$. Observe that $I_{m,0}(a) = \dfrac{a^{m+1}}{m+1}$ for all m. Assume that $n>0$. We have that $I_{m,n}(a) = \int_0^a x^{m}(\ln x)^n dx = \dfrac{x^{m+1}}{m+1} (\ln x)^n|_0^a - \dfrac{n}{m+1} \int_0^a x^{m} (\ln x)^{n-1} dx \Rightarrow I_{m,n}(a) = \dfrac{a^{m+1}}{m+1} (\ln a)^n - \dfrac{n}{m+1} I_{m,n-1} (a)$, since $\lim\limits_{x\to 0} \dfrac{x^{m+1}}{m+1} (\ln x)^n = 0.$
In particular, $\sum_{i=0}^\infty \int_0^a x\dfrac{(x\ln x)^i}{i!}- \dfrac{(x\ln x)^i}{i!}   dx = \sum_{i=0}^\infty I_{i+1,i}(a) - I_{i,i}(a)$.

I'm not sure how to evaluate the latter sum, since I haven't really come up with a good recurrence relation for it.

$\int_0^{a} x(e^{x\ln x} -1) dx = \int_0^a \sum_{i=0}^\infty x(\dfrac{(x\ln x)^i}{i!} - 1) dx  = \sum_{i=0}^\infty \int_0^a x\dfrac{(x\ln x)^i}{i!} - \dfrac{(x\ln x)^i}{i!}   dx.$
Let $f(x)$ be the function $f(x)=x$ or $f(x)\equiv 1$. Since the series $\sum_{i\ge 0} x\dfrac{(x\ln x)^i}{i!}dx$ and $\sum_{i=0}^\infty  \dfrac{(x\ln x)^i}{i!}$ are convergent for any real number x, their difference equals the value of the series $\sum_{i=0}^\infty x(\dfrac{(x\ln x)^i}{i!} - 1)$. Since for $x\in (0,1], \sum_{i=0}^\infty f(x)\dfrac{(-x\ln x)^i}{i!}$ is dominated by the integrable function $f(x)e^{-x\ln x}$, by the Dominated Convergence Theorem, we may interchange the summation and limit in the last equality above. In more detail, for all n, $\sum_{i=0}^n(\dfrac{(x\ln x)^i}{i!}- \dfrac{(x\ln x)^i}{i!} ) \leq \sum_{i=0}^n|\dfrac{(x\ln x)^i}{i!}- \dfrac{(x\ln x)^i}{i!} | \leq \sum_{i\ge 0} |x-1| \dfrac{(-x\ln x)^i}{i!}$, and the latter function is integrable.

A much simpler approach is to use lHospital's rule. Let $n=1/t$ so that $t\to 0^+$ and the expression under limit is $$\frac{1}{t^2}\int_0^t x^{x+1}\,dx$$ Using lHospital's rule we get $$\frac{t^{t+1}}{2t}=\frac{t^t}{2}$$ which tends to $1/2$.
