CONTINUOUS AND DISCONTINUOUS FUNCTIONS

Continuous Functions
Before we talk about Continous functions we will first give a detailed explanation of what a function is.
functions are important in the study of real number system,functions are simply mapping
from one set called the domain to another set called the co-domain.
we can represent so many real life situations as a map or function for example when
every city is mapped to its capital,then an injective mapping will be obtained in that
case.other mathematical examples include the function between the set of natural numbers
$\mathbb{N}$ to the set of real numbers $\mathbb{R}$,
i.e $f:\mathbb{N}\rightarrow\mathbb{R}$
which gives rise to the idea of sequences.so many concept in mathematics deals with the
idea of mapping, from vector spaces,normed linear spaces,metric spaces and many other
spaces in mathematics.other names for functions are mapping,transformations, and morphisms.
Now what is a continuous function or when do we say a function is continuous,using
the cartesian graph, a continuous function is simply a function with no breaking points
on the graph.lets analyse the definition of continuity of functions of the real number
system,now let $f:\mathbb{R}\rightarrow\mathbb{R}$ let $x_0\in\mathbb{R}$, then we say
that $f$ is continuous at $x_0$ if given any ${\epsilon}>0$ then there exists a
${\delta}>0$ such that whenever $|x-x_0|<\delta$ then $|f(x)-f(x_0)|<\epsilon$ in other
words if $x\in(x_0-\delta,x_0+\delta)$ then $f(x)\in(f(x_0)-\epsilon,f(x_0)+\epsilon)$
and that is what a continuous function is.
now lets take an
example.
if given the function $f=x^2$ and we are asked to check the continuity of the function at
$x_0=1$.

Solution:
since the function and point is clearly defined, we simply use the definition
of continuity above to check if the function is continuous at the given point.
let ${\epsilon}>0$ then $|f(x)-f(1)|=|x^2-1^2|$
and we know that $x^2-1=(x+1)(x-1)$ then we can equally replace it.
$|x^2-1|=|x+1||x-1|$, now in order to be able to find $|x+1||x-1|$ to be less than
$\epsilon$ we need to find a ${\delta}>0$  such that  $|x+1||x-1|<{\epsilon}$,
simply let  $|x+1|\leq{m}$ so that $|f(x)-f(1)|\leq{m|x-1|}<\epsilon$ so that m will be
specific, choosing x in (0,2) then $|x|<2$ and $|x+1|\leq{3}$, choose
$\delta=min\{\frac{\epsilon}{3},1\}$ and take an x so that $|x-|<\delta<{1}$ and
$\frac{1}{3}$. now this implies $x\in(0,2)$ and \begin{equation*}
|x+1||x-1|\leq|x-1||x|+1\leq{|x-1|3}<\frac{\epsilon}{3}\times{3}=\epsilon
\end{equation*} and hence our case is proved.

Discontinuous Function
For a function to be discontinuous then it means the function is not continuous and hence
it has breaking points in the cartesian plane.
for $\epsilon>0$ then there exists a $\delta>0$ such that $|x-x_0|<\delta$ does not
imply $|f(x)-f(x_0)|<\epsilon$. in order words for all $x$ satisfying
$x\in(x_0-\delta,x_0+\delta)$ does not imply $f(x)\in(f(x_0)-\epsilon,f(x_0)+\epsilon)$.