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# Use the definition of continuity and the properties of limits to show that the function is continuous at the given number $a$.$p(v) = 2 \sqrt{3v^2 + 1}, \hspace{5mm} a = 1$

## \begin{aligned}\lim _{\boldsymbol{v} \rightarrow 1} p(v) &=\lim _{v \rightarrow 1} 2 \sqrt{3 v^{2}+1}=2 \lim _{v \rightarrow 1} \sqrt{3 v^{2}+1}=2 \sqrt{\lim _{v \rightarrow 1}\left(3 v^{2}+1\right)}=2 \sqrt{3 \lim _{v \rightarrow 1} v^{2}+\lim _{v \rightarrow 1} 1} \\&=2 \sqrt{3(1)^{2}+1}=2 \sqrt{4}=4=p(1)\end{aligned}By the definition of continuity, $p$ is continuous at $a=1$

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

for this problem, we need to show that they function prV which is equal to two squared of three re squared plus one. It's continuous at a equals one. To do this. We recall the definition of continuity at a point a function F of X is continuous at a point X equals E. Whenever it Satisfies the following three conditions, the first one would be for the function to exist at X equals A. That means F. Of a S defined, The 2nd 1 would be for the limit as X approaches a. Of that function to exist and then the last one would be for the limit of this function sx approaches A is equal to the value of the function at A. So if our function is P of V which is equal to two squared of three V squared plus one and is equal to one. And for the first condition we have P of one which is just to square it of three times one squared plus one. That's equal to too scared to four or two times 2 which is four. And since we have a definite value of PF 1 to say that the function is defined at equals one. Next we want to show that the limit exists As X Approaches one. So, and here we have limits As X approaches one of the function two square root of three V squared plus one. This is gonna be um two times the square root of three times one squared plus one. That's the same too, Times Square to four which is also equal to four. And so we have shown that the limit exists and then last these sins. The limit As X approaches one of 2 square it a three V squared plus one. This is equal to four, which is also the value of P at one. Then we have shown that the limit of the function as X approaches one is equal to The value of the function at one. And since the three conditions are met, then by definition the function prvs continuous at A equal to one.

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Limits

Derivatives

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp