Answer
(a) 5
(b)5
(c)5
(d) $f$ is continuous at $x=2$
Work Step by Step
Given $$
f(x)=\left\{\begin{array}{ll}{10 x^{-1}} & {\text { when } x<2} \\ {4 x-3} & {\text { when } x \geq 2}\end{array}\right.
$$
(a) Since
\begin{align*}
\lim _{x \rightarrow2^{-}} f(x)&=\lim _{x \rightarrow2^{-}}10 x^{-1}\\
&=\lim _{x \rightarrow2^{-}}10 (2)^{-1}\\
&=5
\end{align*}
(b) Since
\begin{align*}
\lim _{x \rightarrow2^{+}} f(x)&=\lim _{x \rightarrow2^{+}} (4x-3)\\
&=\lim _{x \rightarrow2^{+}} (8-3)\\
&=5
\end{align*}
(c) $f(2)= 4(2)-3=5$
(d) Since $$\lim _{x \rightarrow2^{+}} f(x)=\lim _{x \rightarrow2^{-}} f(x)=f(2)=5$$
Thus, $f$ is continuous at $x=2$
