use the definition of continuity and the properties of limits to show that the function is continuous at the…

use the definition of continuity and the properties of limits to show that the function is continuous at the given number a.\n\n( p ( v ) = 8 sqrt { 7 v ^ { 2 } + 2 }, quad a = 1 )\n\n( lim _ { v \rightarrow 1 } p ( v ) = lim _ { v \rightarrow 1 } 8 sqrt { 7 v ^ { 2 } + 2 } )\n( = 8 lim _ { v \rightarrow 1 } sqrt { 7 v ^ { 2 } + 2 } ) by the constant multiple law\n( = 8 sqrt { lim _ { v \rightarrow 1 } left( 7 v ^ { 2 } + 2 \right) } ) by the root law\n( = 8 sqrt { lim _ { v \rightarrow 1 } left( 7 v ^ { 2 } \right) + lim _ { v \rightarrow 1 } 2 } ) by the sum law\n( = 8 sqrt { 7 lim _ { v \rightarrow 1 } left( v ^ { 2 } \right) + lim _ { v \rightarrow 1 } 2 } ) by the quotient law\n( = 8 sqrt { 7 cdot ( 1 ) + 2 } ) by the\n( = 24 ) awesome job!\n\nfind ( p ( 1 ) ).\n\n( p ( 1 ) = 24 ) nice job.\n\nthus, by the definition of continuity, ( p ) is contini

use the definition of continuity and the properties of limits to show that the function is continuous at the given number a.\n\n( p ( v ) = 8 sqrt { 7 v ^ { 2 } + 2 }, quad a = 1 )\n\n( lim _ { v \rightarrow 1 } p ( v ) = lim _ { v \rightarrow 1 } 8 sqrt { 7 v ^ { 2 } + 2 } )\n( = 8 lim _ { v \rightarrow 1 } sqrt { 7 v ^ { 2 } + 2 } ) by the constant multiple law\n( = 8 sqrt { lim _ { v \rightarrow 1 } left( 7 v ^ { 2 } + 2 \right) } ) by the root law\n( = 8 sqrt { lim _ { v \rightarrow 1 } left( 7 v ^ { 2 } \right) + lim _ { v \rightarrow 1 } 2 } ) by the sum law\n( = 8 sqrt { 7 lim _ { v \rightarrow 1 } left( v ^ { 2 } \right) + lim _ { v \rightarrow 1 } 2 } ) by the quotient law\n( = 8 sqrt { 7 cdot ( 1 ) + 2 } ) by the\n( = 24 ) awesome job!\n\nfind ( p ( 1 ) ).\n\n( p ( 1 ) = 24 ) nice job.\n\nthus, by the definition of continuity, ( p ) is contini

Answer

Explanation:

Step1: Recall the definition of continuity

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

Step2: Calculate (\lim_{v\rightarrow1}p(v))

[ \begin{align*} \lim_{v\rightarrow1}p(v)&=\lim_{v\rightarrow1}8\sqrt{7v^{2}+2}\ & = 8\lim_{v\rightarrow1}\sqrt{7v^{2}+2}&\text{(Constant Multiple Law)}\ &=8\sqrt{\lim_{v\rightarrow1}(7v^{2}+2)}&\text{(Root Law)}\ &=8\sqrt{\lim_{v\rightarrow1}(7v^{2})+\lim_{v\rightarrow1}2}&\text{(Sum Law)}\ &=8\sqrt{7\lim_{v\rightarrow1}(v^{2})+\lim_{v\rightarrow1}2}&\text{(Constant Multiple Law)}\ &=8\sqrt{7\times(1)^{2}+2}&\text{(Power Law and Direct Substitution Property)}\ &=8\sqrt{7 + 2}\ &=8\times3\ &=24 \end{align*} ]

Step3: Calculate (p(1))

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

Since (\lim_{v\rightarrow1}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\rightarrow1}p(v)=p(1)=24)