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?
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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 $f(x)$. |
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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} |
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