$[\frac{-}{7}$ Points$]$

Use the definition $of$ continuity and the properties $oflimitsto$ show that the function $is$ continuous $at$ the given nur

$,p(v)=4\sqrt[\phantom{2}]{7v^2+2},a=1$

$\underset{v1}{lim}p(v),=\underset{v1}{lim}4\sqrt[\phantom{2}]{7v^2+2},$

$,=4\underset{v1}{lim}\sqrt[\phantom{2}]{7v^2+2},$

$,=4\sqrt[\phantom{2}]{\underset{v1}{lim}(7v^2+2)}by$ the

$,=4\sqrt[\phantom{2}]{\underset{v1}{lim}(7v^2)+\underset{v1}{lim}2}by$ the

$,=4\sqrt[\phantom{2}]{7\underset{v1}{lim}(v^2)+\underset{v1}{lim}2},$

$,=4\sqrt[\phantom{2}]{7*(1)+2}by$ the

$,=,$

Find $p(1).$

$p(1)=$

Thus, $by$ the definition $of$ continuity, $pis$ continuous $ata=1.$