Question

use the definition of continuity and the properties of limits to show that the function is continuous at the given number a.
p(v)=3sqrt3{4v^{2}+5},a = 1
lim_{v
ightarrow1}p(v)=lim_{v
ightarrow1}3sqrt3{4v^{2}+5}=3lim_{v
ightarrow1}sqrt3{4v^{2}+5}=sqrt3{3lim_{v
ightarrow1}(4v^{2}+5)}=sqrt3{lim_{v
ightarrow1}(4v^{2})+lim_{v
ightarrow1}5}=sqrt3{4lim_{v
ightarrow1}(v^{2})+lim_{v
ightarrow1}5}=3sqrt3{4cdot(1)^{2}+5}
find (p(1)).
(p(1)=)

Explanation:

Step1: Recall limit - sum rule

The limit of a sum is the sum of the limits, i.e., $\lim_{v
ightarrow a}(f(v)+g(v))=\lim_{v
ightarrow a}f(v)+\lim_{v
ightarrow a}g(v)$. So, $\lim_{v
ightarrow1}(4v^{2}+ 5)=\lim_{v
ightarrow1}(4v^{2})+\lim_{v
ightarrow1}(5)$.

Step2: Recall constant - multiple rule for limits

The constant - multiple rule for limits states that $\lim_{v
ightarrow a}(cf(v)) = c\lim_{v
ightarrow a}f(v)$, where $c$ is a constant. So, $\lim_{v
ightarrow1}(4v^{2})=4\lim_{v
ightarrow1}(v^{2})$.

Step3: Recall power rule for limits

The power rule for limits states that $\lim_{v
ightarrow a}(v^{n})=a^{n}$ for a positive integer $n$. So, $\lim_{v
ightarrow1}(v^{2}) = 1^{2}=1$. Then $4\lim_{v
ightarrow1}(v^{2})=4\times1 = 4$ and $\lim_{v
ightarrow1}(5)=5$ (since the limit of a constant function $y = c$ as $v
ightarrow a$ is $c$, i.e., $\lim_{v
ightarrow a}c=c$). So, $\lim_{v
ightarrow1}(4v^{2}+5)=4 + 5=9$.

Step4: Recall root - rule for limits

The root - rule for limits states that if $\lim_{v
ightarrow a}f(v)=L$ and $n$ is a positive integer, then $\lim_{v
ightarrow a}\sqrt[n]{f(v)}=\sqrt[n]{L}$ (when $L>0$ for even $n$). Here, $n = 3$ and $f(v)=4v^{2}+5$, $\lim_{v
ightarrow1}(4v^{2}+5)=9$, so $\lim_{v
ightarrow1}\sqrt[3]{4v^{2}+5}=\sqrt[3]{9}$. And since $p(v)=3\sqrt[3]{4v^{2}+5}$, by the constant - multiple rule for limits $\lim_{v
ightarrow1}p(v)=3\lim_{v
ightarrow1}\sqrt[3]{4v^{2}+5}=3\sqrt[3]{9}$.

Step5: Find $p(1)$

Substitute $v = 1$ into $p(v)=3\sqrt[3]{4v^{2}+5}$. We get $p(1)=3\sqrt[3]{4\times1^{2}+5}=3\sqrt[3]{4 + 5}=3\sqrt[3]{9}$.

Since $\lim_{v
ightarrow1}p(v)=p(1)=3\sqrt[3]{9}$, the function $p(v)$ is continuous at $v = 1$.

Answer:

The function $p(v)$ is continuous at $v = 1$ because $\lim_{v
ightarrow1}p(v)=p(1)=3\sqrt[3]{9}$.