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?
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?
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 $f(x)$.
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}
$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{{\rm e}^{#1}}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\pp}{{\cal P}}% \newcommand{\sgn}{{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}$
$$ 0 < \int_{0}^{1}{n\sin\pars{x} \over 1 + n^{2}\,\sqrt{x\,}}\,\dd x < \int_{0}^{1}{n \over 1 + n^{2}\,\sqrt{x\,}\,}\,\dd x = 2n\int_{0}^{1}{x \over 1 + n^{2}x}\,\dd x = {2 \over n}\,\ln\pars{1 + n^{2}} $$
$$ \lim_{n \to \infty}{2 \over n}\,\ln\pars{1 + n^{2}} = 2\lim_{n \to \infty}{2n/\pars{1 + n^{2}} \over 1} = 0 $$
$$ \lim_{n \to \infty}\int_{0}^{1}{n\sin\pars{x} \over 1 + n^{2}\,\sqrt{x\,}}\,\dd x = 0 $$