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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?

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    The function is continuous, so it's integrable. No need for L'Hospital, you aren't calculating limits! – 2017-01-30
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    For the followup question: try $\sin u \leq u$ ($u\geq 0$) and the dominated convergence theorem. – 2017-01-30
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    Isn't the limit $1$? – 2017-01-30
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    @Idonknow pointwise limit, yes. My comment above is to answer the question (which I assume is what the OP is actually asked): compute $\lim_{n\to\infty }\int f_n$ (or, alternatively, find the limit of $(f_n)_n$ in $L_1$). – 2017-01-30

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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

$$\int_{0}^{\pi}\lim_{n\to\infty}f_n = \int_{0}^{\pi} 1=\pi$$