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.

This is very precise, using our fundamental idea of the limit from class 11 we can summarise it as,

A function is continuous at x = c if :

\(\lim\limits_{x \to c}f(x)\) = f(c)

Here we have,

\(f(x) = \begin{cases}x^3-x^2+2x-2&,\quad if\, x ≠1\\4&,\quad if\,x=1\end{cases} \) ….equation 1

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 = 1, 

So we need to check its continuity at x = 1 first.

Clearly,

f(1) = 4 [using eqn 1]

\(\lim\limits_{x \to 1}f(x)\) = \(\lim\limits_{x \to 1}(x^3-x^2+2x-2)\) 

= 1³ - 1² + 2 x 1 - 2 = 0

Clearly,

 \(\lim\limits_{x \to c}f(x)\) ≠ f(c)

∴ f(x) is discontinuous at x = 1. 

Let c be any real number such that c ≠ 0 

f(c) = c³ – c² + 2c – 2 [using eqn 1]

 \(\lim\limits_{x \to c}f(x)\) = \(\lim\limits_{x \to c}(x^3-x^2+2x-2)\) 

= c³ – c² + 2c – 2

Clearly,

 \(\lim\limits_{x \to c}f(x)\) = f(c)

∴ f(x) is continuous for all real x except x = 1