Answer
the function
$$ f(x)=\left\{\begin{array}{ll}{2 x+3,} & {x \leq 4} \\ {7+\frac{16}{x},} & {x>4}\end{array}\right. $$
is continuous on the domain of $f$
Work Step by Step
$$ f(x)=\left\{\begin{array}{ll}{2 x+3,} & {x \leq 4} \\ {7+\frac{16}{x},} & {x>4}\end{array}\right. $$
We can obtain
$$
\lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4^{+}} (7+\frac{16}{x})=11
$$
$$
\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{-}} (2 x+3)=11
$$
We have
$$\lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4^{-}} f(x)$$
Thus
$$ \lim _{x \rightarrow 4} f(x)$$
$$ f(4)= \lim _{x \rightarrow 4} f(x) =11$$
Therefore, the function
$$ f(x)=\left\{\begin{array}{ll}{2 x+3,} & {x \leq 4} \\ {7+\frac{16}{x},} & {x>4}\end{array}\right. $$
is continuous on domain $f$
