Question

Find the points of discontinuity, if any, of the following functions: \[f\left( x \right) = \begin{cases}\frac{x^4 - 16}{x - 2}, & \text{ if } x \neq 2 \\ 16 , & \text{ if }  x = 2\end{cases}\]

Answer 1

Given:  \[f\left( x \right) = \begin{cases}\frac{x^4 - 16}{x - 2}, & \text{ if } x \neq 2 \\ 16 , & \text{ if }  x = 2\end{cases}\]

When x \[\neq\] 2 then 

 \[f\left( x \right) = \frac{x^4 - 16}{x - 2} = \frac{x^4 - 2^4}{x - 2} = \frac{\left( x^2 + 4 \right)\left( x - 2 \right)\left( x + 2 \right)}{x - 2} = \left( x^2 + 4 \right)\left( x + 2 \right)\]

We know that a polynomial function is everywhere continuous.
Therefore, the functions \[\left( x^2 + 4 \right) \text{ and  } \left( x + 2 \right)\] are everywhere continuous.

So, the product function \[\left( x^2 + 4 \right)\left( x + 2 \right)\] is everywhere continuous.

Thus, f(x) is continuous at every x \[\neq\] 2 . 

At x = 2, we have

(LHL at x = 2) =  \[\lim_{x \to 2^-} f\left( x \right) = \lim_{h \to 0} f\left( 2 - h \right) = \lim_{h \to 0} \left[ \left( 2 - h \right)^2 + 4 \right]\left( 2 - h + 2 \right) = 8\left( 4 \right) = 32\]

(RHL at x = 2) =  \[\lim_{x \to 2^+} f\left( x \right) = \lim_{h \to 0} f\left( 2 + h \right) = \lim_{h \to 0} \left[ \left( 2 + h \right)^2 + 4 \right]\left( 2 + h + 2 \right) = 8\left( 4 \right) = 32\]

Also, 

\[f\left( 2 \right) = 16\]

∴ \[\lim_{x \to 2^-} f\left( x \right) = \lim_{x \to 2^+} f\left( x \right) \neq f\left( 2 \right)\]

Thus , 

\[f\left( x \right)\] is discontinuous at x = 2. 

Hence, the only point of discontinuity for

\[f\left( x \right)\]is x = 2.