Determine whether $f(x)={\sin x \over x}$ is uniformly continuous in $\mathbb R$ Determine whether the function$$f(x)={\sin x \over x}$$is uniformly continuous in $\mathbb{R}$.
I am using the definition that for $\epsilon>0$ there exists $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$.
 A: The function $f$ has bounded derivative over $\Bbb R$, given by $$f'(x)=\frac{\cos x}{x}-\frac{\sin{x}}{x^2}$$
It is bounded for $x$ outside a neighborhood of $x=0$. Now note that by definition, the derivative at $0$ is $$\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left( {\frac{{\sin h}}{h} - 1} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sin h - h}}{{{h^2}}} = \mathop {\lim }\limits_{h \to 0} \frac{{\cos h - 1}}{{2h}} = 0$$
by an application of L'Hôpital's rule. Thus $f'(0)=0$. Now, note that again $$\mathop {\lim }\limits_{x \to 0} f'(x) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x\cos x - \sin x}}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{{ - x\sin x}}{{2x}} = 0$$ by L'Hopital. Thus $f'$ is continuous over the real line, and bounded. It follows $f$ is Lipschitz continuous, with constant $M=\sup_{x\in\Bbb R}|f'(x)|$ 
A: Hint:


*

*$\lim_{x \to \pm \infty} \frac{\sin x}{x} =0$;

*$f$ is continuous (I tacitly assume you have defined $f(0)=1$).


Use 1. to split $\mathbb{R} = (-\infty, - M ) \cup [-M,M] \cup (M,+\infty)$, and prove that $f$ is separately UC on each interval. For the middle one, you must know that continuity on a compact set implies uniform continuity.
A: One has
$$f(x):={\sin x\over x}=\int_0^1 \cos(\tau x)\ d\tau\ .$$
It follows that
$$|f(x)-f(y)|\leq \int_0^1|\cos(\tau x)-\cos(\tau y)|\ d\tau\leq \int_0^1\tau |x-y|\ d\tau={1\over2}|x-y|\ .$$
This proves that $f$ is even Lipschitz-continuous.
A: This query has been used as a reference to justify voting to close
a new query.  However, the new query specifically requires that $~\epsilon, \delta~$
be employed.  Therefore, I am filling the gap so that the new
query is answered here.
Define $f(x)$ so that $~f(0) = 1,~~$ and
for all $~\displaystyle x \neq 0, ~~f(x) = \frac{\sin ~x}{x}
~\implies$ 
[since $\lim_{x \to 0} ~f(x) ~= ~1$] $~f(x) ~$ is continuous everywhere.
I will employ an $\epsilon, \delta$ approach to demonstrate that $f$ is
uniformly continuous throughout $\mathbb{R}.$  This answer will then constitute
a stronger result than the OP in the new query is asking for.

To Prove
For any $~\epsilon > 0~$ there exists  $~ \delta > 0 ~$  such that 
$|~f(x) ~-~ f(y) ~| ~<~ \epsilon~$ whenever $~|~x ~-~ y ~| ~<~ \delta$.

Let $~g(x) ~\equiv ~x (\cos x) - (\sin x) ~\implies ~g(0) ~= ~0.$ 
Let $~h(x) ~\equiv ~(\sin x) ~- ~x (\cos x) ~- ~x^2 ~\implies ~h(0) ~= ~0.$ 
For any fixed value of $~x, ~$ and any variable value of $~\delta,$ 
$~~~~~~~~~~~~$let $~J_x(\delta) ~\equiv ~f(x) ~- ~f(x + \delta) ~- ~\delta ~\implies ~~J_x(0) ~= ~0.$ 
Let $~k(x) ~\equiv ~f(0) ~- ~f(x) ~- ~x ~\implies ~k(0) ~= ~0.$ 
Let $~m(x) ~\equiv ~(\sin ~x) ~- ~x ~\implies ~m(0) ~= ~0.$

$\underline{\textbf{Lemma 1:}}$ 
$-\pi/2 ~< ~x ~< ~0 ~\implies ~g(x)  ~> ~0 ~~$ and 
$0  ~< ~x ~< ~\pi/2 ~\implies ~g(x)  ~< ~0.$ 
$\underline{\text{Proof:}}$ 
$g'(x) ~= ~x (-\sin ~x) ~+ ~(\cos ~x) ~- ~(\cos ~x) ~= ~-x (\sin ~x).$ 
Therefore, 
for $~-\pi/2 ~< ~x ~< ~0, ~g(x) ~$ is strictly decreasing $~~$and 
for $~0 ~< ~x ~< ~\pi/2, ~g(x) ~$ is strictly decreasing.
$\underline{\textbf{Lemma 2:}}$ 
For $-\pi/2 ~< ~x ~< ~0, ~f(x) ~$ is strictly increasing $~~$ and 
for $~0 ~< ~x ~< ~\pi/2, ~f(x) ~$ is strictly decreasing. 
$\underline{\text{Proof:}}$ 
For $\displaystyle ~x ~\neq ~0, ~f'(x) ~= 
~\frac{x (\cos ~x) ~- ~(\sin ~x)}{x^2} ~= ~\frac{g(x)}{x^2}.$ 
Invoke Lemma 1.
$\underline{\textbf{Lemma 3:}}$ 
For $0 ~< ~x, ~h(x) ~< ~0 ~~$ and 
for $~x ~< ~0, ~h(x) ~< ~0.$ 
$\underline{\text{Proof:}}$ 
$h'(x) ~= (\cos ~x) ~+ ~x(\sin ~x) ~- ~(\cos ~x) ~- ~2x$ 
$= ~x(\sin ~x) ~- ~2x  ~= x(\sin ~x - 2).$ 
Noting that $~h(0) ~= ~0,$ 
for $~x ~< ~0, ~h(x) ~$ is strictly increasing $~~$ and 
for $~x ~> ~0, ~h(x) ~$ is strictly decreasing.
$\underline{\textbf{Lemma 4:}}$ 
For $~0 ~< ~x, ~0 ~< ~\delta, ~J_x(\delta) < 0.$ 
$\underline{\text{Proof:}}$ 
$\displaystyle ~J_x(\delta) ~= ~\frac{\sin ~x}{x} ~- ~~\frac{\sin ~(x + \delta)}{x + \delta} ~- ~\delta ~\implies$
$\displaystyle 
~\frac{d ~J_x(\delta)}{d ~\delta} ~= 
~- ~\frac{(x + \delta) (\cos[x + \delta]) ~- ~(\sin[x + \delta])}
{(x + \delta)^2} ~- ~1$
$\displaystyle 
~= ~\frac{(\sin[x + \delta]) ~- ~(x + \delta)(\cos[x + \delta]) ~- ~(x + \delta)^2}{(x + \delta)^2}$
$\displaystyle 
~= ~\frac{h(x + \delta)}{(x + \delta)^2}.$
Invoke Lemma 3, noting that $~J_x(0) ~= ~0.$
$\underline{\textbf{Lemma 5:}}$ 
For $~0 ~< ~x ~< \pi/2, ~k(x) < 0.$ 
$\underline{\text{Proof:}}$ 
Since $~f(0) ~= ~1, ~k(x) ~= ~1 ~- ~f(x) ~- ~x.$
By the analysis in the proof to Lemma 2, 
$\displaystyle k'(x) ~= ~- ~\frac{g(x)}{x^2} ~- ~1 ~= 
 ~\frac{h(x)}{x^2}.$
By Lemma 3,
this implies that for $0 ~< ~x ~< \pi/2, ~k'(x) < 0.$ 
Noting that $~k(x) ~= ~0, ~$ the Lemma is proven.
$\underline{\textbf{Lemma 6:}}$ 
For $~0 ~< ~x, ~m(x) < 0.$ 
$\underline{\text{Proof:}}$ 
$m'(x) ~= ~(\cos ~x) ~- ~1 ~\implies$ 
for $~x ~< ~\pi/2,  ~m'(x)~$ is strictly negative. 
Thereafter, $~m'(x)~$ is never $~> ~0.$ 
Therefore, with $~m(0) = 0, ~m(x)$ 
is strictly decreasing on $~(0, ~\pi/2)~$ and 
thereafter, is strictly non-increasing.
$\underline{\text{Identity 7:}}$ 
For $~0 ~< ~|a - b| ~< ~(\pi/4),$
$\displaystyle |(\sin ~a) ~- ~(\sin ~b)| 
~= ~2\left|\left(\sin\frac{a - b}{2}\right) ~\left(\cos\frac{a + b}{2}\right)\right|$
$\displaystyle \leq ~2\left|\left(\sin\frac{a - b}{2}\right) \right|$
$\displaystyle = ~2 \times \left(\sin ~\left|\frac{a - b}{2}\right|\right)
~< ~\text{[by Lemma 6]} ~|a - b|.$

WLOG $~x ~< ~y.$ 
Set $\displaystyle ~\delta = \min\left[\left(\frac{\pi}{4} \times \epsilon/2\right), ~\left(\pi/4\right)\right] 
~\implies \frac{\delta}{\pi/4} ~< ~(\epsilon/2).$
Note that $~\forall ~x, ~f(x) ~= ~f(-x). ~$ Assume that $~|x - y| ~< ~\delta.$
$\underline{\text{Case 1:} ~~x ~< ~0 ~\leq ~y}$ 
$|f(x) - f(y)| ~\leq ~|f(x) - f(0)|  ~+ |f(0) - f(y)|~$ 
$= ~|f(-x) - f(0)| ~+ ~|f(0) - f(y)|.$ 
By Lemma 5, each of the two terms in the line above is $~< ~\delta.$ 
Therefore, their sum is less than $~\epsilon.$
$\underline{\text{Case 2:} ~~x ~= ~0 ~< ~y}$ 
By Lemma 5, $~|f(x) - f(y)| ~< ~\delta ~< ~\epsilon.$
$\underline{\text{Case 3:} ~~0 ~< ~x ~< ~\pi/4 ~~~~ 
\text{and} ~~~~x ~< ~y}$ 
Let $~\alpha ~= ~(y - x) ~\implies ~\alpha ~< ~\delta.$ 
Then, by Lemma 4, 
$~|f(x) - f(y)| ~= ~J_x(\alpha) ~+ ~\alpha ~< ~\alpha ~< ~\delta ~< ~\epsilon.$
$\underline{\text{Case 4:} ~~\pi/4 ~\leq ~x ~< ~y}$ 
Let $~\alpha ~= ~(y - x) ~\implies ~\alpha ~< ~\delta.$
$\displaystyle |f(x) - f(y)| ~= ~\left|\frac{\sin ~x}{x} ~- ~\frac{\sin ~y}{y}\right|$
$\displaystyle = ~\left|\frac{\sin ~x}{x} ~- \frac{\sin ~x}{y} ~+ \frac{\sin ~x}{y} ~- ~\frac{\sin ~y}{y}\right|$
$\displaystyle \leq ~\left|\frac{\sin ~x}{x} ~- \frac{\sin ~x}{y}\right| ~+ ~\left|\frac{\sin ~x}{y} ~- ~\frac{\sin ~y}{y}\right|.$
Using Lemma 6,
$\displaystyle ~ \left|\frac{\sin ~x}{x} ~- \frac{\sin ~x}{y}\right| ~= \left(\frac{1}{x} - \frac{1}{y}\right) ~|\sin ~x|$
$\displaystyle < ~\left(\frac{y - x}{xy}\right) \times (x) ~= ~\frac{\alpha}{y} ~< ~\frac{\delta}{\pi/4} ~< ~\frac{\epsilon}{2}.$
Using Identy 7,
$\displaystyle \left|\frac{\sin ~x}{y} ~- ~\frac{\sin ~y}{y}\right| ~= ~\left(\frac{1}{y}\right) \times \left|\sin ~x ~- ~\sin ~y\right|$
$\displaystyle < ~\left(\frac{1}{\pi/4}\right) \times (\alpha) ~< ~\frac{\delta}{\pi/4} < \frac{\epsilon}{2}.$
Thus, $~|f(x) - f(y)| ~< ~\epsilon.$
$\underline{\text{Case 5:} ~~x ~< ~y ~\leq ~0}$ 
Since $~f(x) = f(-x), ~|f(x) - f(y)| ~= ~|f(-x) - f(-y)| .$ 
Therefore, Case 5 is proven by Cases 2 through 4, collectively.
