f(x) ={[(x^4-16)/(x-2),if x ≠ 2][16,if x = 2]

Question

Find the points of discontinuity, if any, of the following functions :

\(f(x) = \begin{cases}\frac{x^4-16}{x-2}&,\quad if\, x ≠2\\16&,\quad if\,x=2\end{cases} \)

Answer 1

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,

…Equation 1

Note : 

[for changing the expression used identity :– (a²–b²) = (a+b)(a–b)] 

Note : x – 2 is cancelled from numerator and denominator only because x ≠ 2, else we can’t cancel them 

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 = 2, 

So we need to check its continuity at x = 2 first. 

Clearly, 

f(2) = 16 [using eqn 1]