## Calculus, 10th Edition (Anton)

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 $f$
$$f(x)=\left\{\begin{array}{ll}{2 x+3,} & {x \leq 4} \\ {7+\frac{16}{x},} & {x>4}\end{array}\right.$$ 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 $f$