Prove that $\lim_{n\to\infty} n^2 \int _0^{1/n}{x^{x+1}}dx = {1\over2}$. Prove that
$$\lim_{n\to\infty} n^2 \int _0^{1/n}{x^{x+1}}dx = {1\over2}.$$
My Attempt:
When I saw the integral it was tempting to see the integrand as  $x^x x$ and to think of $lim_{x\to0}$ $x^x$ which when evaluated using L-Hopital's gives 1, will it be helpful in reaching a desired conclusion?
 A: $\newcommand{\d}{\,\mathrm{d}}$If you enforce $x=t/n$, you get: $$\lim_{n\to\infty}n^2\int_0^{1/n}x^{x+1}\d x=\lim_{n\to\infty}\int_0^1t^{1+t/n}n^{-t/n}\d t$$We have: $$0\le t^{1+t/n}n^{-t/n}\le1$$The dominated convergence theorem is then applicable: It is also the case that $1\ge n^{-t/n}\ge n^{-1/n}\to1$ as $n\to\infty$, so in pointwise evaluation of the integrand (justified by DCT) we find: $$\lim_{n\to\infty}\int_0^1t^{1+t/n}n^{t/n}\d t=\int_0^1 t^1\d t=\frac{1}{2}$$
EDIT: A much cleaner way to prove this directly without appellation to the DCT is as follows: $$t\ge t^{1+t/n}n^{-t/n}\ge t^{1+1/n}n^{-1/n}$$On the desired interval, so we have: $$\begin{align}\frac{1}{2}&=n^{1/n}\int_0^1t\d t\ge\int_0^1t^{1+t/n}n^{-t/n}\d t\ge n^{-1/n}\int_0^1t^{1+1/n}\d t\\&= \frac{1}{2+1/n}\cdot n^{-1/n}\end{align}$$From which the squeeze theorem makes an obvious limit of $1/2$.
A: From $\frac{x^{x+1}}{x}=x^x=e^{x\ln(x)}$ and $\lim_{x\to 0^+}x\ln(x)=0$ (prove this yourself), it can be seen that $\lim_{x\to 0^+}\frac{x^{x+1}}{x}=1$. It follows that for $\varepsilon>0$ fixed arbitrarily, we can find a $\delta>0$ satisfying
$$\left|\frac{x^{x+1}}{x}-1\right|<2\varepsilon\text{ whenever }0<x<\delta$$
which says the same thing as
$$\begin{equation}(1-2\varepsilon)x<x^{x+1}<(1+2\varepsilon)x\text{ whenever }0<x<\delta\end{equation}\tag{$\ast$}$$
Pick $N\in\mathbb N$ sufficiently large to get $\frac{1}{N}\leq\delta$. Then for any $n\in\mathbb N$ with $n>N$, we have $0<\frac{1}{n}<\delta$. Integrating the inequality $(\ast)$ over $[0,1/n]$ gives
$$\int_0^{\frac{1}{n}}(1-2\varepsilon)xdx<\int_0^{\frac{1}{n}}x^{x+1}dx<\int_0^{\frac{1}{n}}(1+2\varepsilon)xdx\text{ for every }n>N$$
and consequently, after evaluating the left and rightmost integrals,
$$\left(\frac{1}{2}-\varepsilon\right)\frac{1}{n^2}<\int_0^{\frac{1}{n}}x^{x+1}dx<\left(\frac{1}{2}+\varepsilon\right)\frac{1}{n^2}\text{ for every }n>N$$
Multiplying through by $n^2$ and then subtracting $1/2$ yields
$$-\varepsilon<n^2\int_0^{\frac{1}{n}}x^{x+1}dx-\frac{1}{2}<\varepsilon\text{ for every }n>N$$
or
$$\left|n^2\int_0^{\frac{1}{n}}x^{x+1}dx-\frac{1}{2}\right|<\varepsilon\text{ for every }n>N$$
Since $\varepsilon>0$ was arbitrary, it follows that
$$\lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}x^{x+1}dx=\frac{1}{2}$$
A: Since you already recived good answers for the limit, let us try to have asymptotics.
We have, by expansion around $x=0$,
$$x^{x+1}=\sum_{k=1}^\infty \frac 1{(k-1)!}x^k\,\big[\log (x)\big]^{k-1} $$ Now, consider the integrals
$$I_k=\int_0^{\frac 1{n}}x^k\,\big[\log (x)\big]^{k-1}\,dx$$
$$I_1=\frac{1}{2 n^2}\qquad I_2=-\frac{3 \log (n)+1}{9 n^3}\qquad I_3=\frac{8 \log ^2(n)+4 \log (n)+1}{32 n^4}$$
$$n^2 \big[I_1+ I_2+\frac 12 I_3+\cdots\big]=\frac 12-\frac{3 \log (n)+1}{9 n}+\frac{8 \log ^2(n)+4 \log (n)+1}{64 n^2}+\cdots$$
Trying for $n=10$, the above gives
$$\frac{28169-1884 \log (10)+72 \log ^2(10)}{57600}=0.42036$$ while numerical integration would give $0.41985$
A: $$\lim_{n\to\infty} n^2 \int _0^{1/n}{x^{x+1}}dx = \lim_{n\to\infty}\frac{ \int _0^{1/n}{x^{x+1}}dx}{\frac{1}{n^2}}$$
Using L’Hôpital’s and Leibniz’s rules,
$$\lim_{n\to\infty}\frac{(\frac {1}{n})^{\frac{n+1}{n}}(\frac{-1}{n^2})}{\frac{-2}{n^3}}=\frac12 \lim_{n\to\infty}\left(\frac{1}{n}\right)^{\left(\tfrac{1}{n}\right)}=\frac12.$$
, using the limit the OP has already mentioned in the question.
A: If integral limits gap approaches zero, integral is simply area of trapezoid.
Let $ε = 1/n$
$\displaystyle
\lim_{ε\to {0^+}} \frac{1}{ε^2} \int _0^ε{x^{x+1}}dx
= \lim_{ε\to {0^+}} \frac{1}{ε^2} \left( \frac{ε^{ε+1}+0^1}{2} \right) ε
= \lim_{ε\to {0^+}} \frac{ε^ε}{2} 
= \frac{1}{2}
$
