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 ) = 8 sqrt { 7 v ^ { 2 } + 2 }, quad a = 1 )

( lim _ { v
ightarrow 1 } p ( v ) = lim _ { v
ightarrow 1 } 8 sqrt { 7 v ^ { 2 } + 2 } )
( = 8 lim _ { v
ightarrow 1 } sqrt { 7 v ^ { 2 } + 2 } ) by the constant multiple law
( = 8 sqrt { lim _ { v
ightarrow 1 } left( 7 v ^ { 2 } + 2
ight) } ) by the root law
( = 8 sqrt { lim _ { v
ightarrow 1 } left( 7 v ^ { 2 }
ight) + lim _ { v
ightarrow 1 } 2 } ) by the sum law
( = 8 sqrt { 7 lim _ { v
ightarrow 1 } left( v ^ { 2 }
ight) + lim _ { v
ightarrow 1 } 2 } ) by the quotient law
( = 8 sqrt { 7 cdot ( 1 ) + 2 } ) by the
( = 24 ) awesome job!

find ( p ( 1 ) ).

( p ( 1 ) = 24 ) nice job.

thus, by the definition of continuity, ( p ) is contini

Explanation:

Step1: Recall the definition of continuity

A function \(f(x)\) is continuous at \(x = a\) if \(\lim_{x
ightarrow a}f(x)=f(a)\)

Step2: Calculate \(\lim_{v

ightarrow1}p(v)\)

$$
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$$

Step3: Calculate \(p(1)\)

$$ p(1)=8\sqrt{7\times(1)^{2}+2}=8\sqrt{7 + 2}=24 $$

Since \(\lim_{v
ightarrow1}p(v)=p(1) = 24\), by the definition of continuity, the function \(p(v)\) is continuous at \(a = 1\)

Answer:

The function \(p(v)\) is continuous at \(a = 1\) because \(\lim_{v
ightarrow1}p(v)=p(1)=24\)