Question

Find the points of discontinuity, if any, of the following functions:  $$f\left(x\right)=\{\begin{array}{lll}-2,& \text{ if }& x\le -1\\ 2x,& \text{ if }& -1<x<1\\ 2,& \text{ if }& x\ge 1\end{array}$$

Answer 1

The given function f is $$f\left(x\right)=\{\begin{array}{lll}-2,& \text{ if }& x\le -1\\ 2x,& \text{ if }& -1<x<1\\ 2,& \text{ if }& x\ge 1\end{array}$$ 

The given function is defined at all points of the real line.

Let c be a point on the real line.

Case I:

${\displaystyle \text{If}c<-1\text{then}f\left(c\right)=-2\text{and}\underset{x\to c}{lim}\left(x\right)=\underset{x\to c}{lim}(-2)=-2}$

${\displaystyle \therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)}$

Therefore, f is continuous at all points x, such that x < −1

Case II:

${\displaystyle \text{If c =1 then} f\left(c\right)=f(-1)=-2}$

The left hand limit of f at x = −1 is,

${\displaystyle \underset{x\to -1}{lim}f\left(x\right)=\underset{x\to -1}{lim}f(-2)=-2}$

The right hand limit of f at x = −1 is,

${\displaystyle \underset{x\to -1}{lim}f\left(x\right)=\underset{x\to -1}{lim}f\left(2x\right)=2\times (-1)=-2}$

${\displaystyle \therefore \underset{x\to -1}{lim}f\left(x\right)=f(-1)}$

Therefore, f is continuous at x = −1

Case III:

${\displaystyle \text{if -1 < c < 1,then}f\left(c\right)=2c}$

${\displaystyle \underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}f\left(2x\right)=2c}$

${\displaystyle \therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)}$

Therefore, f is continuous at all points of the interval (−1, 1).

Case IV:

${\displaystyle \text{if c = 1,then} f\left(c\right)=f\left(1\right)=2\times 1=2}$

The left hand limit of f at x = 1 is,

${\displaystyle \underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}2=2}$

The right hand limit of f at x = 1 is,

${\displaystyle \underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}2=2}$

${\displaystyle \therefore \underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}f\left(c\right)}$

Therefore, f is continuous at x = 2

Case V:

${\displaystyle \text{If c > 1 , then f(c)=2 and} \underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}\left(2\right)=2}$

${\displaystyle \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)}$

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.