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Determine $$\lim_{n\to \infty}\int_{[0,1]}\frac{n\sin{x}}{1+n^2\sqrt{x}}dx$$

I know the three convergence theorems, but to no avail:

$1.$ Monotone Convergence:

The series $f_{n}(x):=\frac{n\sin{x}}{1+n^2\sqrt{x}}$ is not monotonic increasing on $[0,1]$, so the conditions are not met.

$2.$ Fatou:

Note: $\liminf_{n\to \infty}\int_{[0,1]}\frac{n\sin{x}}{1+n^2\sqrt{x}}dx\geq\int_{[0,1]}\liminf_{n\to \infty}\frac{n\sin{x}}{1+n^2\sqrt{x}}dx=0$

Which aids us no further.

$3.$ Dominated Convergence Theorem:

The only function $h$ to fulfill $|f_{n}|\leq h, \forall n \in \mathbb N$ that comes to mind is $|\frac{n\sin{x}}{1+n^2\sqrt{x}}|=\frac{n\sin{x}}{n^2\sqrt{x}}=\frac{\sin{x}}{n\sqrt{x}}\leq \frac{\sin{x}}{\sqrt{x}}=:h(x).$

But how do I show $h$ is $\in \mathcal{L}^1$?

Is there anything I am missing? Any guidance is greatly appreciated.

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Use dominated convergence with $h(x)=\frac{\sin(x)}{x}$ for $x\in(0,1]$ and $h(0)=1$.

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You’re almost there... the map $h : x \mapsto \frac{\sin x}{\sqrt x}$ has $0$ for limit as $x \to 0$. Hence can be extended by continuity on $[0,1]$ and is integrable on that interval.

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