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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 $

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

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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.

The University of Alabama