Uniform Continuity of functions

[Bling Fizikst]

Homework Statement

below

Relevant Equations

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Discuss uniform continuity of the following functions:
$\tan x$ in $[0,\frac{\pi}{2})$
$\frac{1}{x}\sin^2 x$ in $(0,\pi]$
$\frac{1}{x-3}$ in $(0,3),(4,\infty),(3,\infty)$

I am completely new to this uniform continuity and couldn't find a lot of examples to learn the solving pattern .
For the first problem , i tried to use the definition ,
for $x,y$ in the given interval , $|x-y|<\delta\implies |\tan x-\tan y|=\frac{|\sin(x-y)|}{|\cos x\cos y|}\leq \frac{1}{|\cos x\cos y|}$
Just stuck in this and the other problems .

 

[Hill]

If you are stuck trying to show uniform continuity of the function, try to show that the function is NOT uniformly continuous.

 

[fresh_42]

Continuity is a local phenomenon. It always means continuous at a point $x_0.$ The usual definition is that for every $\varepsilon >0$ there is a $\delta >0$ such that $|f(x)-f(x_0)|<\varepsilon$ whenever $|x-x_0|<\delta.$

If you look closer to that definition, then you will find out that $x_0$ is chosen prior to $\delta$ as is $\varepsilon.$ This means that it actually should have been written as:
$$
\forall\;\varepsilon>0 \;\;\exists \;\delta(x_0,\varepsilon) \, : \, |x-x_0|<\delta(x_0,\varepsilon) \Longrightarrow |f(x)-f(x_0)|<\varepsilon\;.
$$
The difference is, that I noted that $\delta=\delta(x_0,\varepsilon)$ depends on location $x_0$ as well as on the margin $\varepsilon.$ We only say $x\to f(x)$ is continuous without mentioning the location if $f$ is continuous for all $x_0$ in the domain of $f.$ But $\delta$ may depend on the various locations $x_0.$

A third definition is uniform continuity. It simply expresses the fact that we can chose $\delta = \delta(\varepsilon)$ without using the location $x_0.$ We have
$$
\forall\;\varepsilon>0 \;\;\exists \;\delta(\varepsilon) \, : \, |x-x_0|<\delta(\varepsilon) \Longrightarrow |f(x)-f(x_0)|<\varepsilon\;.
$$
for any two $x,x_0$ in the domain of $f.$

The question in the exercise is thus: Can $\delta$ be chosen independently from the location $x_0$? Can we choose $\delta=\delta(\varepsilon)$ only depending on our choice of the margin $\varepsilon$ anywhere in the domain or not?

The typical example is $f(x)=\dfrac{1}{x}$ on $(0,1].$ It is continuous but not uniformly continuous because our $\delta$ would be
$$
\delta=\delta(x_0,\varepsilon)= \min\left\{\dfrac{x_0}{2}\, , \,\dfrac{x_0^2\varepsilon}{2}\right\}
$$
as
$$\left|f(x)-f(x_0)\right|=\left|\dfrac{1}{x}-\dfrac{1}{x_0}\right|=\left|\dfrac{x-x_0}{xx_0}\right|\leq \dfrac{2\left|x-x_0\right|}{x_0^2}<\dfrac{2\delta}{x_0^2}<\varepsilon\;.$$