## Calculus 8th Edition

$p$ is continuous at $a=1$.
A function $f$ is continuous at a number $a$ if $\displaystyle \lim_{x\rightarrow a}f(x)=f(a)$ ------------- $\displaystyle \lim_{v\rightarrow 1}p(v)=\lim_{v\rightarrow 1}2\sqrt{3v^{2}+1}\quad$...The limit of a constant times a function $=2\displaystyle \lim_{v\rightarrow 1}\sqrt{3v^{2}+1}$ ... $\displaystyle \lim_{x\rightarrow a}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x\rightarrow a}f(x)}$ where $n$ is a positive integer $=2\sqrt{\lim_{v\rightarrow 1}(3v^{2}+1)} \quad$...The limit of a sum $=2\sqrt{\lim_{v\rightarrow 1}3v^{2}+\lim_{v\rightarrow 1}1}\quad$...The limit of a constant times a function $=2\sqrt{3\lim_{v\rightarrow 1}v^{2}+\lim_{v\rightarrow 1}1}\quad$...evaluate $=2\sqrt{3(1)^{2}+1}$ $=2\sqrt{4}$ $=4$ $p(1)=2\sqrt{3\cdot 1^{2}+1}=2\cdot\sqrt{4}=4= \displaystyle \lim_{v\rightarrow 1}p(v)$ By the definition, $p$ is continuous at $a=1$.