Uniform Convergence $f_n(x)=\sin\left(\frac{x}{n}\right)$ on $[-R, R]$

Question

I wish to show uniform convergence of $f_n(x)=\sin\left(\frac{x}{n}\right)$ to $g(x)=0$ for all $x \in [-R, R]$, where $R$ is a positive real number.

To do this, we need:

$\forall \epsilon>0, \forall x\in [-R, R], \exists N\in \mathbb N$ such that $n>N$ means $|f_n(x)-g(x)| < \epsilon$.

So we are interested in estimating:

$$|\sin(x/n)|$$

The familiar problem, how do we pick $N$? For $n$ sufficiently large we see that since $x$ is bounded by $R$ and finite we know $x/n$ can be made as small as we like, at which point $\sin(x/n)$ behaves like $x/n$ which tends to $0$. How do I make this more precise?

Answer 1

A classical estimate is $|\sin t|\le |t|$ for all $t\in\Bbb R$. So $$|f_n(x)|=|\sin(x/n)|\le |x/n|\le R/n$$ for all $x\in[-R,R]$. So Given $\varepsilon>0$, let $N\in\Bbb N$ such that $\frac{R}{N}<\varepsilon$. So for $n\ge N$ and $x\in[-R,R]$ $$|f_n(x)|\le \frac{R}{n}<\varepsilon$$ and you're done.

Answer 2

hint

By MVT, For $ X>0 $, $$|\sin(X)-\sin(0)|=|X\cos(c)|\le |X|$$ For $ X<0 $, $$|\sin(-X)|\le |-X|$$

So,

$$(\forall X\in \Bbb R)\;\; |\sin(X)|\le |X|$$

to get

$$(\forall x\in[-R,R])\; \left|\sin\left(\frac xn\right)\right|\le \left|\frac xn\right|\le \frac Rn$$

and $$\sup_{[-R,R]}|f_n-g|\le \frac Rn$$

with $$\lim_{n\to+\infty}\frac Rn=0$$