an example of convergence theorems I have this question and I am not able to solve it. 

$\ f_n(x)= \frac{{n}^{3/2}x}{1+n^2x^2}$ , $x\in[0,1]$, $n\in N$
Using convergence theorems, show that $\displaystyle \lim_{n \to
 \infty} \int_0^1 f_n(x)dx\, =0$

I tried to find a dominated integrable function but I could not find it. This sequence is not increasing. So I could not use the monotone convergence theorem. I also tried to use the derivative of arctan(nx). How can I proceed? Can someone help me? Thank you. 
 A: I don't know if this is what you'd like to see, but here is my approach:
$$\lim\limits_{n \to \infty} \int\limits_{x=0}^{1} \frac{n^{3/2}x}{1+n^2x^2} \mathrm{d}x=\lim\limits_{n \to \infty} \int\limits_{x=0}^{1} \frac{n^{3/2}}{2n^2}\frac{2n^{2}x}{1+n^2x^2} \mathrm{d}x=\lim\limits_{n \to \infty} \frac{n^{3/2}}{2n^2} \left. \log(1+n^2x^2)\right|_{x=0}^{1}=\lim\limits_{n \to \infty} \frac{n^{3/2}}{2n^2} \left(\log(1+n^2)-\log(1)\right)=\lim\limits_{n \to \infty} \frac{n^{3/2}}{2n^2}\log(1+n^2)=\lim\limits_{n \to \infty} \frac{\log(n^2+1)}{2\sqrt{n}}=\frac{1}{2}\lim\limits_{n \to \infty} \frac{\log(n^2+1)}{\sqrt{n}}=\frac{1}{2}\cdot0=0$$
Since $\log(n^2+1)$ grows asymptotically slower than $\sqrt{n}$ as $n$ approaches $\infty$.
A: By AM-GM: 
$$\begin{align}\dfrac{1+n^2x^2}{n^{3/2}x^{3/2}} &= \dfrac{1}{(nx)^{3/2}} + (nx)^{1/2} \\&= \dfrac{1}{(nx)^{3/2}} + \dfrac{(nx)^{1/2}}{3} + \dfrac{(nx)^{1/2}}{3} + \dfrac{(nx)^{1/2}}{3} \\
&\ge 4 \sqrt[4]{\dfrac{1}{(nx)^{3/2}}\cdot \dfrac{(nx)^{1/2}}{3}\cdot\dfrac{(nx)^{1/2}}{3}\cdot\dfrac{(nx)^{1/2}}{3}} \\&= \dfrac{4}{3^{3/4}}.\end{align}$$ 
Hence, $\dfrac{n^{3/2}x^{3/2}}{1+n^2x^2} \le \dfrac{3^{3/4}}{4}$, and thus, $\dfrac{n^{3/2}x}{1+n^2x^2} \le \dfrac{3^{3/4}}{4x^{1/2}}$, which is integrable on $[0,1]$.
You can then apply Lebesgue's dominated convergence theorem in the usual way.
