Calculate $\lim_{n\to\infty}\int_0^1\frac{nx}{1+n^2 x^3}\,dx$. I need calculate $\displaystyle\lim_{n\to\infty}\int_0^1\frac{nx}{1+n^2 x^3}\,dx$.
I showed that $f_n(x)=\frac{nx}{1+n^2 x^3}$ converges pointwise to $f\equiv 0$ but does not converge uniformly in [0,1]. 
Also $f_n$ has a maximum when $\displaystyle x=\frac{1}{\sqrt[3]{2n^2}}$.
By the last thing, I can't use the theorem of integration of uniform convergence.
I wait that you can give me a hint for this exercise. Thanks!
 A: By $AM-GM$, 
$$2n\sqrt{x^3}=2\sqrt{n^2x^3} \le 1+n^2x^3.$$
So then you can bound the integrand by 
$$\frac{nx}{1+n^2x^3} \le \frac{nx}{2nx^{3/2}} =\frac{1}{2\sqrt{x}}$$
which is integrable on $[0,1]$, and then apply dominated convergence.
A: Let $t=nx$ and then
\begin{eqnarray}
I_n&=&\int_0^1\frac{nx}{1+n^2 x^3}\,dx\\
&=&\int_0^n\frac{t}{1+n^2 \frac{t^3}{n^3}}\frac{1}{n}\,dt\\
&=&\int_0^n\frac{t}{n+t^3}dt\\
&=&\int_0^\infty\chi_{[0,n]}(t)\frac{t}{n+t^3}dt\\
\end{eqnarray}
Noting
$$ \chi_{[0,n]}(t)\frac{t}{n+t^3}\le\frac{t}{1+t^3}, \int_0^\infty\frac{t}{1+t^3}dt<\infty, \lim_{n\to\infty}\chi_{[0,n]}(t)\frac{t}{n+t^3}=0 $$
by the DCT, one has
$$ \lim_{n\to\infty}I_n=0. $$
A: With $t:=n^{2/3}x$,$$\lim_{n\to\infty}\int_0^1\frac{nx}{1+n^2 x^3}\,dx=\lim_{n\to\infty}n^{-1/3}\int_0^{n^{2/3}}\frac t{1+t^3}\,dt$$
and the expression is asymptotic to 
$$\frac{2\pi}{3\sqrt3}n^{-1/3}.$$
(Definite integral obtained with Wolfram Alpha, but the exact value is inessential.)

We can be more accurate by evaluating the asymptotics of the integral,
$$\int_0^{n^{2/3}}\frac t{1+t^3}dt=\int_0^\infty\frac t{1+t^3}dt-\int_{n^{2/3}}^\infty\frac t{1+t^3}dt\approx\frac{2\pi}{3\sqrt3}-n^{-2/3},$$
as for large $t$, $\dfrac t{1+t^3}\approx\dfrac1{t^2}$.
Hence,
$$L_n\sim\frac{2\pi}{3\sqrt3}n^{-1/3}-\frac1n.$$
