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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 $

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03:10

Daniel Jaimes

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 5

Continuity

Limits

Derivatives

Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

Lectures

04:40

In mathematics, the limit of a function is the value that the function gets very close to as the input approaches some value. Thus, it is referred to as the function value or output value.

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

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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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