Continuity and integration using Dominated Convergence theorem I have to use the Dominated Convergence Theorem to show that $\lim \limits_{n \to \infty}$ $\int_0^1f_n(x)dx=0$ where $f_n(x)=\frac{n\sqrt{x}}{1+n^2x^2}$.  
I did the following:
$$\frac{n\sqrt{x}}{1+n^2x^2} <\frac{n\sqrt{x}}{n^2x^2} = \frac{x^{-\frac{3}{2}}}{n}\leq x^{-\frac{3}{2}} $$
But $$\int_0^1x^{-\frac{3}{2}}dx=\frac{-2}{\sqrt{x}}\biggr|_0^1$$ which doesn't seem right. Any help will be appreciated. 
 A: Use for example:
$$\frac{n\sqrt x}{1+n^2x^2}=\frac{1}{\sqrt x}\cdot\frac{1}{\frac{1}{nx}+nx}\le \frac{1}{2\sqrt x}$$
since $u+\frac{1}{u}\ge 2$ for $u>0$
And $\int_0^1\frac{1}{2\sqrt x}=1<\infty$
(this is equivalent to AM-GM inequality on the denominator $1+n^2x^2\ge2nx$)
A: A worse bound, a slightly different approach:
$$ \frac{n\sqrt x}{1+n^2x^2} \le \frac{n\sqrt x}{\sqrt{1+n^2x^2}} = n \sqrt{ \frac{x}{1+n^2x^2}} = \frac{1}{\sqrt{x}} \sqrt{\frac{1}{1+\frac{1}{n^2x^2}}} \le \frac{1}{\sqrt{x}} $$
Where we used tha facts that $x \ge \sqrt{x} $ if $x \ge 1$ and $\sqrt{\frac{1}{1+x}} \le 1 $ on $[0,+\infty)$.
A: Let $f_n(x)=\frac{n\sqrt{x}}{1+n^2x^2}$.  Then, differentiating with respect to $n$ reveals
$$\begin{align}
\frac{df_n(x)}{dn}&=\frac{\sqrt{x}}{1+n^2x^2}-\frac{2n^2x^{5/2}}{(1+n^2x^2)^2}\\\\
&=\frac{\sqrt{x}(1-n^2x^2)}{(1+n^2x^2)^2}\tag1
\end{align}$$
From $(1)$, it is easy to see that $\frac{df_n(x)}{dn}=0$ when $n=1/x$.  Moreover, we can see from $(1)$ that $f_n(x)$ is a maximum when $n=1/x$.  Therefore, we have
$$\sup_{n} \left(\frac{n\sqrt{x}}{1+n^2x^2}\right)=\frac{1}{2\sqrt{x}}$$
whereby we have a dominating function that is integrable on $[0,1]$.
