What is $\lim_{n \to \infty} \space n^2\int_{0}^{1/n} x^{x+1} dx$? How do we evaluate 
$$\lim_{n \to \infty} \space n^2\int_{0}^{1/n} x^{x+1} dx\quad ?$$ 
I know that 
$$\lim_{z \to 0+} \space \dfrac{\int_{0}^z x^{x+1} dx}{z^2}=\dfrac12,$$
and I think the asked limit should also be $1/2$ but I can not prove it. 
Please help.  
 A: Since
$$
\color{#C00000}{\lim_{x\to0}x^x=1}
$$
and
$$
\color{#00A000}{n^2\int_0^{1/n}x\,\mathrm{d}x=\frac12}
$$
we have
$$
\begin{align}
&\color{#C00000}{1}\cdot\color{#00A000}{\frac12}\\
&=\color{#C00000}{\lim_{n\to\infty}\left[\inf_{x\in[0,1/n]}x^x\right]}\color{#00A000}{\lim_{n\to\infty}\left[n^2\int_0^{1/n}x\,\mathrm{d}x\right]}\\
&\le\lim_{n\to\infty}\left[n^2\int_0^{1/n}x^{x+1}\,\mathrm{d}x\right]\\
&\le\color{#C00000}{\lim_{n\to\infty}\left[\sup_{x\in[0,1/n]}x^x\right]}\color{#00A000}{\lim_{n\to\infty}\left[n^2\int_0^{1/n}x\,\mathrm{d}x\right]}\\
&=\color{#C00000}{1}\cdot\color{#00A000}{\frac12}
\end{align}
$$
By the Squeeze Theorem, we have
$$
\lim_{n\to\infty}\left[n^2\int_0^{1/n}x^{x+1}\,\mathrm{d}x\right]=\frac12
$$
A: let $F(x)=\int x^{x+1}dx$ then $$\lim\limits_{n\to\infty}n^2\int\limits_0^{\frac{1}{n}}x^{x+1}dx=\lim\limits_{n\to\infty}n^2\left(F\left(\frac{1}{n}\right)-F(0)\right)=\lim\limits_{n\to\infty}n^2F\left(\frac{1}{n}\right)=\infty.0.$$ Then we get 
$$\lim\limits_{n\to\infty}\frac{F\left(\frac{1}{n}\right)}{\frac{1}{n^2}}=\frac{0}{0}$$ Then using L'Hospital's rule we have $$\lim\limits_{n\to\infty}\frac{f\left(\frac{1}{n}\right).\left(\frac{-1}{n^2}\right)}{\frac{-2}{n^3}}=\lim\limits_{n\to\infty}\frac{f\left(\frac{1}{n}\right).n}{2}=\lim\limits_{n\to\infty}\frac{1}{2\sqrt[n]n}=\frac{1}{2}$$ where $f(x)=x^{x+1}.$
A: If you develop $x^{x+1}$ as a Taylor series built around $x=0$, you obtain $$x+x^2 \log (x)+O\left(x^3\right)$$ and the integral between the given bounds is then $$\frac{1}{2} \left(\frac{1}{n}\right)^2+\left(\frac{1}{n}\right)^3 \left(\frac{1}{3}
   \log
   \left(\frac{1}{n}\right)-\frac{1}{9}\right)+O\left(\left(\frac{1}{n}\right)^4\right)$$ So, the limit is $\frac{1}{2}$.
