The function f is defined in $[–5,5]$ as $f\left(x\right) = x$, if $x$ is rational and $f\left(x\right) = –x$, if $x$ is irrational. Then

Let a is a rational number other than 0, in [-5, 5], then 
f(a)=a and $\underset{x\to a}{\lim }f\left(x\right)=-a$
[As in the immediate neighbourhood of a rational number, we find irrational numbers] 
∴f(x) is not continuous at any rational number 

If a is irrational number, then f(a)=−a and$\underset{x\to a}{\lim }f\left(x\right)=a$
∴f(x) is not continuous at any irrational number 

clearly 
$\underset{x\to 0}{\lim }f\left(x\right)$=f(0)=0 
∴f(x) is continuous at x = 0

$\text{ L.H.L. }=\underset{x\to a}{lim} f\left(x\right)=\underset{h\to 0}{lim} f(a-h)=0\text{ or }1$

$\text{ R.H.L. }=\underset{x\to a+}{lim} f\left(x\right)=\underset{h\to 0}{lim} f(a+h)=0\text{ or }1\text{, }$

So, $f\left(x\right)$ is discontinuous at $x=a a\ne 0$.

$\text{ For }a=0\Rightarrow f\left(0\right)=0\text{ R.H.L. }=\underset{x\to 0+}{lim} f\left(x\right)=0$

Similarly $L.H.L.= 0$. Thus $f\left(x\right)$ is continuous at $x = 0$ only.