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Explain why the function is discontinuous at the given number $ a $. Sketch the graph of the function.

$ f(x) = \frac{1}{x + 2} \hspace{55mm} a = -2 $

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$f(x)=\frac{1}{x+2}$ is discontinuous at $a=-2$ because $f(-2)$ is undefined

01:36

Daniel Jaimes

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 5

Continuity

Limits

Derivatives

Campbell University

Oregon State University

Harvey Mudd College

Boston College

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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$17-22$ Explain why the fu…

Okay, let's explain why the function F of X equals one over X plus two is discontinuous at a equals negative two. So let's let's talk about limits. The limit as X approaches to From the left of one over X plus two. This number is going to be something like a little bit less than two, right? So like 1.999999999. It helps to think of it that way. Mhm. Sorry, this is negative two. So this should be negative 1.9999999 -1.99999. So if we plug that in, we get the limit. Okay, We already floated in so we don't need the limit sign. We get one over and this is sort of theoretical, so you don't really want to show this work if you can avoid it, but negative one point 99999999999 plus two is going to be something 0000011 over An extremely small number leads us to infinity. Now let's look at the other side, negative two minus. This is going to be something like negative two point oh one of one over X plus two. So that gives us one over. If we plug that in -0.000001 that points us to negative infinity. If you're kind of wondering about this, think of one one over 0.1 Which is equal to one over 1/10. You get 10 from that if you have one over a very very very small number, that leads you to a bigger number when you divide it. So that's why this is discontinuous because the limit from the left, uh huh Of Alfa Becks is not equal to the limit from the right. So therefore the limit as X approaches negative to one over X plus two does not exist. Do you two? A vertical. This continuity which is an asem tope. So let's just draw this really quickly just to make sure that we've got this down. This is -2, as we've said, um as we approach this way, so if we have negative two over on the left hand side, it's going to go down to negative infinity on the right hand side. If we have a little bit more we see up here this part, then we're going to have positive infinity, Which means there's an assumed toe right here at -2. So that is why this function is discontinuous. The limit from the left is not equal the limit from the right, they are equal to infinity. And that indicates a vertical discontinuity. The vertical ascent toe. That's why it's discontinuous.

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