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: The key fact here is that
$$\lim_{x \rightarrow 0+} x^x = 1$$
There are various ways to show the above. As a result,
for any $1 > \epsilon > 0$, there is an $N$ such that
for $n > N$, on $(0,1/n]$ one has
$$(1 - \epsilon)x < x^{x + 1} < (1 + \epsilon) x$$
Insert this into your integral and let $n \rightarrow \infty$...
A: 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 $1/2$.
A: First of all, it is worth attempting to convert the sequence of integrals on varying intervals to one in which the occurring integrals refer to a fixed interval. In our situation this can be simply achieved by the change of variable $x=\frac{y}{n}$, mapping the interval $[0, 1] \ni y$ on $\left[0, \frac{1}{n}\right] \ni x$. Explicitly we have:
$$\begin{align}
n^2\int_0^{\frac{1}{n}}x^{x+1}\mathrm{d}x=n^2\int_0^1 \left(\frac{y}{n}\right)^{\frac{y}{n}+1}\frac{1}{n}\mathrm{d}y=\int_0^1 y\left(\frac{y}{n}\right)^{\frac{y}{n}}\mathrm{d}x.
\end{align}$$
By any of the various means available, one can establish the strictly decreasing monotony of the map $\left[0, \frac{1}{e}\right] \to [0, 1]$ given by $x \mapsto x^x$. From this one infers that for any $n \in \mathbb{N}$ with $n \geqslant 3$ and any $x \in [0, 1]$ the relation:
$$0 \leqslant 1-\left(\frac{x}{n}\right)^{\frac{x}{n}}\leqslant 1-\left(\frac{1}{n}\right)^{\frac{1}{n}}=1-\frac{1}{\sqrt[n]{n}}$$
takes place, grounds on which one can furthermore easily show that the sequence of functions $[0, 1] \to \mathbb{R}$ with the $n$-th term given by $x \mapsto x\left(\frac{x}{n}\right)^{\frac{x}{n}}$ converges uniformly to the inclusion $x \mapsto x$ of $[0, 1] \to \mathbb{R}$. At this point one no longer even needs to worry about subtleties of Lebesgue integration but can simply turn to the elementary properties of Riemann integration to conclude that:
$$n^2\int_0^{\frac{1}{n}}x^{x+1}\mathrm{d}x \xrightarrow{n \to \infty} \int_0^1x\mathrm{d}x=\frac{1}{2}.$$
A: Another way is as follow: knowing that $$\int{x^{x+1}}dx=\frac{x^2}{2}+A_3x^3+A_4x^4+\cdots+Constant=F(x)$$ where the $A_i$'s are complicated expressions given by, for example Wolfram, in the attached figure of which we only make explicit  $A_3=\dfrac{3\ln(x)-1}{9}$. We have so $$\int_0^{\frac 1n}x^{x+1}dx=\frac{1}{2n^2}+A_3\frac {1}{n^3}+A_4\frac{1}{n^4}+\cdots$$ Thus the required limit is clearly equal to $\dfrac12$

A: Using
$$x^{x+1}=x \,e^{x\log(x)}=\sum_{k=0}^\infty \frac 1{k!}\,x^{k+1} \log^k(x)$$ and integrating termwise
$$ \int_0^{\frac 1 n} x^{x+1} \,dx=\frac{1}{2 n^2}-\frac{3 \log (n)+1}{9 n^3}+\frac{8 \log ^2(n)+4 \log
   (n)+1}{64 n^4}+\cdots$$ shows the limit and how it is approached.
