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

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.

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