Lebesgue Dominated Convergence example How do I use Lebesgue Dominated Convergence Theorem to evaluate 
$$\lim_{n \to \infty}\int_{[0,1]}\frac{n\sin(x)}{1+n^2\sqrt x}dx$$
What dominating function to use here?
 A: Consider this for a solution (it does not use dominated convergence). In $[0,1]$, we have the following inequality
\begin{equation}
0 \leq \frac{n \sin(x)}{1+ n^2 \sqrt{x}} < \frac{n}{n^2 \sqrt{x}} = \frac{1}{n \sqrt{x}}
\end{equation}
Taking integral over $[0,1]$ on both sides, we get
\begin{equation}
\int_{[0,1]} \frac{n \sin(x)}{1+ n^2 \sqrt{x}} < \int_{[0,1]} \frac{1}{n \sqrt{x}} = \left[ \frac{2 \sqrt{x}}{n} \right]_0^1
\end{equation}
taking limit $n \to \infty$ on both sides, you can show that 
\begin{equation}
\lim_{n \to \infty} \int_{[0,1]} \frac{n \sin(x)}{1+ n^2 \sqrt{x}} = 0
\end{equation}
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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$$
0 < \int_{0}^{1}{n\sin\pars{x} \over 1 + n^{2}\,\sqrt{x\,}}\,\dd x
<
\int_{0}^{1}{n \over 1 + n^{2}\,\sqrt{x\,}\,}\,\dd x
=
2n\int_{0}^{1}{x \over 1 + n^{2}x}\,\dd x
=
{2 \over n}\,\ln\pars{1 + n^{2}}
$$


$$
\lim_{n \to \infty}{2 \over n}\,\ln\pars{1 + n^{2}}
=
2\lim_{n \to \infty}{2n/\pars{1 + n^{2}} \over 1} = 0
$$

$$
\lim_{n \to \infty}\int_{0}^{1}{n\sin\pars{x} \over 1 + n^{2}\,\sqrt{x\,}}\,\dd x
=
0
$$
A: Note that for $x\in(0,1]$ and $n\geq 1$, we have:
$$\frac{n\sin(x)}{1+n^2\sqrt{x}} \leq \frac{n}{1+n^2\sqrt{x}} \leq \frac{n}{n^2\sqrt{x}} \leq \frac{1}{n\sqrt{x}} \leq \frac{1}{\sqrt{x}}.$$
Thus, if $g(x) = \frac{1}{\sqrt{x}}$ for $0\lt x\leq 1$ and $g(x)=0$ if $x=0$, then $0\leq f_n(x)\leq g(x)$ almost everywhere in $[0,1]$, where $f_n(x) = \frac{n\sin(x)}{1+n^2\sqrt{x}}$. Since
$$\int g(x)\,d\mu = \int_{[0,1]}\frac{1}{\sqrt{x}}\,d\mu= 2\sqrt{x}\Biggm|_0^1 = 2\lt \infty$$
by Lebesgue's Dominated Convergence Theorem you know that if $f(x)$ equals the pointwise limit of the $f_n(x)$ almost everywhere on $[0,1]$, then 
$$\lim_{n\to\infty}\int_{[0,1]}\frac{n\sin(x)}{1+n^2\sqrt{x}}\,d\mu = \int_{[0,1]}f(x)\,d\mu.$$
So to use Dominated Convergence you now need to figure out an $f(x)$. 
