Question

How do you determine whether the discontinuity is removable or non removable? Show with an example.

Answer 1

Removal discontinuity and non removable discontinuity exist in a rational function.

Removable Discontinuity:

Suppose we have a rational function $f(x)$ which is discontinuous at some point say $x = a$ but if the limit $\lim_{x \to a } f(x)$ exists then we can say that the function has a removable discontinuity at $x = a$

Example:

Let us assume that

$f(x) = \frac{x^2 - 9}{x+3}$

Observing the function, we can say that the function has discontinuity at $x = -3$ but the limit of the function at $x \to -3$

is given by

$\lim_{x \to -3} \frac{x^2 - 9}{x + 3} \\ = \lim_{x \to -3} \frac{(x+3)(x-3)}{x+3} \hspace{1 cm}$ [ $\because a^2 - b^2 = (a+b)(a-b)$ ]

$= \lim_{x \to -3} (x-3) \hspace{1 cm}$ [ By canceling common factors $(x+3)$ from both numerator and denominator ]

$= -3 - 3 \hspace{1 cm}$ [ By substituting $x = - 3$ into the limit function ]

$= -6$

Hence, the limit of the function exists and $x = - 3$ is the removable discontinuity of the function $f(x)$

$\\$

Non Removable Discontinuity:

In non removable discontinuity, after canceling the common factors from both numerator and denominator, the function still has a discontinuity at some point.

Example:

Let us assume that

$g(x) = \frac{x^2 - x - 2}{x^2 - 5x - 6}$

$\hspace{0.5 cm} \,\, \,\,= \frac{x^2 - 2x + x - 2}{x^2 - 6x + x - 6 }$

$\hspace{0.5 cm} \,\,\,\, = \frac{x(x-2) + 1(x-2)}{x(x-6) +1(x-6)}$

$\hspace{0.5 cm} \,\,\,\, = \frac{(x-2)(x+1)}{(x-6)(x+1)}$

$\hspace{0.5 cm} \,\,\,\, = \frac{x-2}{x-6}$

After canceling the common factors $(x+1)$ from both numerator and denominator, we can see that the function $g(x)$ still has discontinuity at $x = 6 .$ This is non removable discontinuity.