Question

Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}\frac{x^4 + x^3 + 2 x^2}{\tan^{- 1} x}, & \text{ if } x \neq 0 \\ 10 , & \text{ if }  x = 0\end{cases}\]

Answer 1

 When x \[\neq\] 0, then

\[f\left( x \right) = \frac{x^4 + x^3 + 2 x^2}{\tan^{- 1} x}\]

We know that

\[x^4 + x^3 + 2 x^2\] is a polynomial function which is everywhere continuous.
Also,

\[\tan^{- 1} x\] is everywhere continuous.
So, the quotient function

\[\frac{x^4 + x^3 + 2 x^2}{\tan^{- 1} x}\] is continuous at each x \[\neq\] 0.

Let us consider the point x = 0.

Given: 

\[f\left( x \right) = \binom{\frac{x^4 + x^3 + 2 x^2}{\tan^{- 1} x}, \text{ if }  x \neq 0}{10, \text{ if  } x = 0 }\]

We have
(LHL at x = 0) = \[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} \left( \frac{\left( - h \right)^4 + \left( - h \right)^3 + 2 \left( - h \right)^2}{\tan^{- 1} \left( - h \right)} \right) = \lim_{h \to 0} \left( \frac{\left( h \right)^3 - \left( h \right)^2 + 2\left( h \right)}{- \frac{\tan^{- 1} \left( h \right)}{h}} \right) = \frac{0}{\left( - 1 \right)} = 0\]

(RHL at x = 0) = \[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right) = \lim_{h \to 0} \left( \frac{\left( h \right)^4 + \left( h \right)^3 + 2 \left( h \right)^2}{\tan^{- 1} \left( h \right)} \right) = \lim_{h \to 0} \left( \frac{\left( h \right)^3 + \left( h \right)^2 + 2\left( h \right)}{\frac{\tan^{- 1} \left( h \right)}{h}} \right) = \frac{0}{1} = 0\]

Also,

\[f\left( 0 \right) = 10\]

∴ \[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right) \neq f\left( 0 \right)\]

Thus, 

\[f\left( x \right)\]  is discontinuous at x = 0.

Hence, the only point of discontinuity for 

\[f\left( x \right)\]  is x = 0.