Uniform continuity of $\frac{\sin(x)}{x}$

Question

Let $f=\frac{\sin(x)}{x}$, then $f$ is continuous on $\mathbb{R}$. To show this function is uniformly continuous, we can subdivide its domain into $[-a,a]$ (such that $0\in [a,b]$) and $\mathbb{R}\backslash [-a,a]$. With the second piece of the domain, it is simple to show that $f$ is uniformly continuous by, for example, finding that $\forall \varepsilon>0$, $\delta = \varepsilon a^2 -a $, $\forall x\in \mathbb{R}\backslash[-a,a]$.

But with the second part, since $[-a,a]$ contains $\{0\}$, it doesn't seem possible to show uniform continuity using inequalities. So we can probably show that $f$ is Lipschitz on $[-a,a]$ by showing that its derivative is bounded. However, it doesn't seem to be bounded. What else can be done?

score 1 · Answer 1 · answered Nov 18 '16 at 08:05

1

Hint: Note that the function $$\left\{\begin{array}{c l}\frac{\sin x}{x}, & x\neq 0 \\ 1, & x=0 \end{array}\right.$$ for $x\in[-a,a]$ is continuous, and $[-a,a]$ is a closed interval.

answered Nov 18 '16 at 08:05

detnvvp

8,357

How can you show that $f$ is uniformly continuous over $\Bbb R$? – Learnmore Nov 18 '16 at 08:28
$[-a,a]$ is a compact set, and $f$ is continuous on $[-a,a]$. Does this mean that $f$ is uniformly continuous on $[-a,a]$? – sequence Nov 18 '16 at 08:39
Yes, this is correct. – detnvvp Nov 18 '16 at 20:34

Uniform continuity of $\frac{\sin(x)}{x}$

1 Answers1