Recently, I have come across a difficult exercise. How to prove that the function $$ f_n: (0, \pi] \to \mathbb{R},\quad f_n(x) = \frac{n\sin(\frac{x}{n})}{x}$$ is lebesgue-integrable for all $n \in \mathbb{N}$?
I tried to apply Hospital's rule but did not succeed.
EDIT:
I missed another part. How to decide whether it is still integrable for $\lim_{n\to\infty} f_n$ and how to calculate this limit?
Note that for each $x>0$ we have
$$\sin(x) < x$$
and so $\sin(\frac{x}{n}) < \frac{x}{n}$ for any $x>0$, $n>0$.
Now if you write
$$f_n(x)=\frac{\sin(\frac{x}{n})}{\frac{x}{n}}$$
then it is clear that $f_n(x) \leq 1$ for every $n, x$. Also on $(0, \pi]$ each $f_n$ is bounded from below by $0$. Actually $f_n(x) \geq -1$ on any interval. Anyway $\lVert f_n(x)\rVert\leq 1$ for any $n, x$.
Thus dominated convergence theorem applies and the integral of the limit exists and is equal to the integral of pointwise limit. Now since
$$\lim_{n\to\infty}\frac{\sin(\frac{x}{n})}{\frac{x}{n}}=\lim_{z\to 0}\frac{\sin(z)}{z}=1$$
then the integral of the limit is equal to
$$\int_{0}^{\pi}\lim_{n\to\infty}f_n = \int_{0}^{\pi} 1=\pi$$