Question

Discuss continuity of $f\left(x\right)=\frac{x^3-64}{\sqrt{x^2+9-5}}$ for $x\ne 4$
$=10$ at $x=4$

Answer 1

$f\left(x\right)=\{\begin{array}{c}\frac{x^3-64}{\sqrt{x^2+9-5}},x\ne 4\\ 10,x=0\end{array}$

Checking continuity at $x=4,\Rightarrow$

$\underset{x\to 4^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{{(4-h)}^3-64}{\sqrt{{(4-h)}^2+4}}=0$

$\underset{x\to 4^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{{(4+h)}^3-64}{\sqrt{{(4+h)}^2+4}}=0$

$\Rightarrow f\left(4\right)=10$

$\Rightarrow \underset{x\to 4^{-}}{lim}f\left(x\right)=\underset{x\to 4^{+}}{lim}f\left(x\right)\ne f\left(4\right)$

$\Rightarrow$The function is 'isolated point discontinuous' as $\underset{x\to 4}{lim}f\left(x\right)$ exist and $f\left(4\right)$ also exist but they are not equal.