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) = 9\sqrt{7v^2 + 2}, \\ a = 1
\lim\limits_{v\to 1} p(v) = \lim\limits_{v\to 1} 9\sqrt{7v^2 + 2}
\\ \\ \\ \\ \\ = 9 \lim\limits_{v\to 1} \sqrt{7v^2 + 2} \\ \\ \\ \\ \\ \\ \\ \\ \\ by the \underline{\quad\quad select\quad\quad}
\\ \\ \\ \\ \\ = 9\sqrt{\lim\limits_{v\to 1} \left(7v^2 + 2\
ight)} \\ \\ \\ \\ \\ \\ \\ by the \underline{\quad\quad select\quad\quad}
\\ \\ \\ \\ \\ = 9\sqrt{\lim\limits_{v\to 1} \left(7v^2\
ight) + \lim\limits_{v\to 1} 2} \\ by the \underline{\quad\quad select\quad\quad}
\\ \\ \\ \\ \\ = 9\sqrt{7 \lim\limits_{v\to 1} \left(v^2\
ight) + \lim\limits_{v\to 1} 2} \\ by the \underline{\quad\quad select\quad\quad}
\\ \\ \\ \\ \\ = 9\sqrt{7 \cdot (1) + 2} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ by the \underline{\quad\quad select\quad\quad}
\\ \\ \\ \\ \\ = \boxed{\quad\quad}

find p(1).
p(1) = \boxed{\quad\quad}

thus, by the definition of continuity, p is continuous at a = 1.

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Explanation:

Step1: Calculate the limit step by step

First, we know that for the limit of a constant multiple, we can use the constant multiple rule. Then for the square root function, we use the root rule for limits. For the sum of functions, we use the sum rule, and for the constant multiple of a function, we use the constant multiple rule again. Finally, we substitute the value of \( v = 1 \) for the limit of a power function (since \( \lim_{v
ightarrow1}v^{2}=1^{2} = 1 \)) and the limit of a constant ( \( \lim_{v
ightarrow1}2=2 \) ).

We have:
\( \lim_{v
ightarrow1}p(v)=9\sqrt{7\cdot(1)^{2}+2} \)
First, calculate the expression inside the square root: \( 7\cdot(1)^{2}+2=7 + 2=9 \)
Then, \( \sqrt{9} = 3 \)
Then, \( 9\times3=27 \)

Step2: Calculate \( p(1) \)

Substitute \( v = 1 \) into the function \( p(v)=9\sqrt{7v^{2}+2} \)
\( p(1)=9\sqrt{7\times(1)^{2}+2}=9\sqrt{7 + 2}=9\sqrt{9}=9\times3 = 27 \)

Answer:

For the limit calculation, the result is \( 27 \). For \( p(1) \), the result is also \( 27 \). So the filled - in values are \( 27 \) (for the limit) and \( 27 \) (for \( p(1) \)).