Locate the points of discontinuity, if any, for the function $f\left(x\right)= \begin{cases} \dfrac{x^3-64}{x^2-16}, & x \neq 4 \\ 1, & x=4 \end{cases}$
$\begin{aligned}
f\left(x\right) & = \begin{cases}
\dfrac{x^3-64}{x^2-16}, & x \neq 4 \\
1, & x=4
\end{cases} \\
& \\
& = \begin{cases}
\dfrac{\left(x-4\right)\left(x^2+4x+16\right)}{\left(x+4\right)\left(x-4\right)}, & x \neq 4 \\
1, & x=4
\end{cases} \\
& \\
& = \begin{cases}
\dfrac{x^2+4x+16}{x+4}, & x \neq 4 \\
1, & x=4
\end{cases} \\
& \\
& = \begin{cases}
\dfrac{x^2+4x+16}{x+4}, & x>4 \\
\dfrac{x^2+4x+16}{x+4}, & x<4 \\
1, & x=4
\end{cases}
\end{aligned}$
When $x>4$, $f\left(x\right)=\dfrac{x^2+4x+16}{x+4}$ is continuous (polynomial function).
When $x<4$, $f\left(x\right)=\dfrac{x^2+4x+16}{x+4}$ is continuous (polynomial function).
When $x=4$,
$\text{Left Hand Limit (LHL)} = \lim\limits_{x \to 4^-}f\left(x\right)=\lim\limits_{x \to 4^-} \; \dfrac{x^2+4x+16}{x+4}$
Let $x=4-h$
As $x \to 4^-, \; h \to 0$
$\begin{aligned}
\therefore \text{LHL} & =\lim\limits_{h \to 0} \; \dfrac{\left(4-h\right)^2+4\left(4-h\right)+16}{4-h+4} \\
& = \lim\limits_{h \to 0} \; \dfrac{16-8h+h^2+16-4h+16}{8-h} \\
& = \lim\limits_{h \to 0} \; \dfrac{h^2-12h+48}{8-h} \\
& = \dfrac{0-0+48}{8-0} = 6
\end{aligned}$
$\text{Right Hand Limit (RHL)} = \lim\limits_{x \to 4^+}f\left(x\right)=\lim\limits_{x \to 4^+} \; \dfrac{x^2+4x+16}{x+4}$
Let $x=4+h$
As $x \to 4^+, \; h \to 0$
$\begin{aligned}
\therefore \text{RHL} & =\lim\limits_{h \to 0} \; \dfrac{\left(4+h\right)^2+4\left(4+h\right)+16}{4+h+4} \\
& = \lim\limits_{h \to 0} \; \dfrac{16+8h+h^2+16+4h+16}{8+h} \\
& = \lim\limits_{h \to 0} \; \dfrac{h^2+12h+48}{8+h} \\
& = \dfrac{0+0+48}{8+0} = 6
\end{aligned}$
$f\left(4\right)=1$
$\therefore$ $\text{LHL}=\text{RHL}\neq f\left(4\right)$
$\implies$ $x=4$ is a point of discontinuity.