<s>[INST] Find $\lim_{x\to0}\sum_{n=1}^{\infty}\frac{\sin x}{4+n^2x^2}$ I have a problem with finding the following limit. I suspect that it should be easy, but really I don't have a clue.
$$
\lim_{x\to0}\sum_{n=1}^{\infty}\frac{\sin x}{4+n^2x^2}
$$
 [/INST] Note that if $(n-1)x \leqslant u \leqslant nx \leqslant v \leqslant (n+1)x,$ then
$$\frac{1}{4 + v^2} \leqslant  \frac{1}{4 +n^2x^2} \leqslant \frac{1}{4 + u^2}, $$
and
$$\int_{nx}^{(n+1)x}\frac{dv}{4 + v^2} \leqslant  \frac{x}{4 +n^2x^2} \leqslant \int_{(n-1)x}^{nx}\frac{du}{4 + u^2}, $$
Summing we get
$$\int_x^\infty \frac{dv}{4 + v^2} \leqslant \sum_{n=1}^\infty \frac{x}{4 +n^2x^2} \leqslant \int_0^\infty \frac{du}{4 + u^2} .$$
Using the squeeze principle,
$$\lim_{x \to 0}\sum_{n=1}^\infty \frac{x}{4 +n^2x^2} = \int_0^\infty \frac{du}{4 + u^2}.$$
Hence,
$$\lim_{x \to 0}\sum_{n=1}^\infty \frac{\sin x}{4 +n^2x^2} =  \lim_{x \to 0} \frac{\sin x}{x} \sum_{n=1}^\infty \frac {x}{4 +n^2x^2} \\ = \int_0^\infty \frac{du}{4 + u^2}$$
</s>