How do you prove that $\frac{\sin x}{n \sin (\frac{x}{n})}$ converges uniformly?

Question

For $n \in \mathbb N$, define $$f_n(x) = \frac{\sin x}{n \sin (\frac{x}{n})}.$$ As $n \to \infty$, I have shown that $\{f_n\}$ converges (pointwise) to $\frac{\sin x}{x}$. I am trying to evaluate $\lim\limits_{n \to \infty} \int_0^\pi f_n(x) \, dx$, and my attempt is to show the uniform convergence of the integrand on $[0,\pi]$, so that I can interchange the operations of limit and integral to obtain $\lim\limits_{n \to \infty} \int_0^\pi f_n(x) \, dx = \int_0^\pi \frac{\sin x}{x} \, dx$.

By plotting $f_n(x)$ with a graphing calculator and letting $n \to \infty$, I am quite convinced that $\{f_n\}$ indeed converges uniformly to $\frac{\sin x}{x}$. Nevertheless, I can't seem to prove this proposition, whether by the definition of uniform convergence or by Cauchy criterion. Perhaps my difficulty stems from the fact that the bounds on $f_n$ depends on the interaction between $\sin x$, $\sin(\frac{x}{n})$, and $n$. That is, finding a bound on one or two expressions can't help in establishing a bound for the whole function. So far I have tried applying the inequalities $0 < \sin x \leq 1$ and $\sin x \leq x$ for $0 < x < \pi$, but without any luck.

Any idea on how to prove this statement? Or is there a theorem on uniform convergence that is applicable for this case?

(Note: This question is related to the Dirichlet kernel in Fourier analysis, which can be expressed as $D_N(x) = \frac{1}{2\pi} + \frac{1}{\pi} \sum_{n=1}^N \cos nx = \frac{1}{2\pi} \frac{\sin(N+\frac{1}{2})x}{\sin (\frac{x}{2})}$. However, I don't think this is relevant at this point.)

Answer 1

Hint: Choose $m$ such that $n >m$ implies $|1-\frac {\sin (x/n)} {x/n}| <\epsilon$ for all $x \in [0,\pi]$. Then $\frac {\sin x} {(1+\epsilon)x}<f_n(x)<\frac {\sin x} {(1-\epsilon)x}$ for all $x$ if $n >m$. Can you finish?

[Check that $|f_n(x)-\frac {\sin x} x| <\frac {\epsilon} {1-\epsilon} $ for $n >m$ using the fact that $0 \leq \frac {\sin x} x \leq 1$].

Answer 2

Since for $x\ge 0$: $$\sin x \ge x - \frac{x^3}{6}$$ we have $$\begin{align} \|f_n(x) - f(x)\| &= \left\|\frac{\sin x}{n\sin(x/n)} - \frac{\sin x}{x}\right\| \\ &= \color{blue}{\left\|\sin x\right\|}\left\| \frac{1}{n\color{red}{\sin(x/n)}} - \frac{1}{x}\right\| \\ &\le \color{blue}1\cdot \left\|\frac{1}{n\color{red}{\left(\frac{x}{n} - \frac{x^3}{6n^3}\right)}} - \frac{1}{x}\right\| \\ &= \left\|\frac{1}{x - \frac{x^3}{6n^2}} - \frac{1}{x}\right\| \\[1mm] &\xrightarrow{n\to\infty}0 \end{align}$$