Answer
$p$ is continuous at $a=1$.
Work Step by Step
A function $f$ is continuous at a number $a$ if
$\displaystyle \lim_{x\rightarrow a}f(x)=f(a)$
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$\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 $a=1$.
