<s>[INST] 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. 
 [/INST] 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 $[0,1]$.
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