A real function f is said to be continuous at x = c, 

Where c is any point in the domain of f 

If :

\(\lim\limits_{h \to 0}f(c-h)\) = \(\lim\limits_{h \to 0}f(c+h)\) = f(c)

where h is a very small ‘+ve’ no.

i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

A function is continuous at x = c if :

\(\lim\limits_{x \to c}f(x)\) = f(c)

Here we have,

\(f(x) = \begin{cases}\frac{x}{|x|}&,\quad x ≠0\\0&,\quad x=0\end{cases} \) …….equation 1

The function is defined for all real numbers, so we need to comment about its continuity for all numbers in its domain 

(domain = set of numbers for which f is defined)

Function is changing its nature (or expression) at x = 0, 

So we need to check its continuity at x = 0 first.

NOTE : 

Definition of mod function : 

|x| = \(\begin{cases}-x,x <0\\x,x≥0\end{cases} \)

 LHL = \(\lim\limits_{h \to 0}f(0-h)\) 

= \(\lim\limits_{h \to 0}f(-h)\) 

= \(\frac{-h}{-(-h)}\)

= \(\frac{-h}{h}\) 

= -1

[using eqn1 and idea of mod fn]

f(0) = 0 

[using eqn 1] 

Clearly,

LHL ≠ RHL ≠ f(0) 

∴ function is discontinuous at x = 0 

Let c be any real number such that c > 0

∴ f(c) = \(\frac{c}{|c|}\)

= \(\frac{c}{c}\) = 1

[using eqn 1]

And,

\(\lim\limits_{x \to c}f(x)\) = \(\lim\limits_{x \to c}\frac{c}{|c|}\) 

= \(\lim\limits_{x \to c}\frac{c}{c}\) = 1

Thus,

\(\lim\limits_{x \to c}f(x)\) = f(c)

∴ f(x) is continuous everywhere for x > 0. 

Let c be any real number such that c < 0

∴ f(c) = \(\frac{c}{|c|}\) = \(\frac{c}{-c}\) = -1

[using eqn 1 and idea of mod fn]

And,

 \(\lim\limits_{x \to c}f(x)\) = \(\lim\limits_{x \to c}\frac{c}{|c|}\) 

= \(\lim\limits_{x \to c}\frac{c}{-c}\) = -1

Thus,

 \(\lim\limits_{x \to c}f(x)\) = f(c)

∴ f(x) is continuous everywhere for x < 0.

Hence, 

We can conclude by stating that f(x) is continuous for all Real numbers except zero.