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# Find the limit, if it exists. If the limit does not exist, explain why. $\displaystyle \lim_{x \to -2}\frac{2 - |x|}{2 + x}$

## 1

Limits

Derivatives

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

This is problem number forty four of the Stuart Calculus. Eighth edition, section two point three. Final amid. If it exists, it's the limit does not exist. Explain why Here we have a limited six bridges. Negative too. Of the function to minus work. The absolute value backs divide where the quantity two plus sex. Okay, since this is a complete limit, not just a limit from the left or the right, we wantto determine first. Whether approach in a tutu makes this an indeterminate form or an undefined value. Two minus the absolutely of negative too is to minus two two plus native who? Zero. So we have a zero division here, divide by zero. Add some *** too. So we know that the value or the function is on to find a native to so we were unable to determine that it exists are at this point. Um, but we can still prove that ism it exists as long as the limited left negative two equals limit from the right. I'm negative too. On that the equal each other. So what we want to do is we want to first deal with this absolute value function, and we want to rewrite this Lim. I'm using what we know about the absolute value function of X. We know that when X is greater than zero or equal to zero, it's equal the function X. However, when X is negative, he also value is meant to change the sign and so dysfunctions represented by a negative X and for our interests since the limit of expertise and negative too. This is on ly in this region. So we will not be considering the absolute value function too. Take this form because we are only concerned with the X approaches negative to region. So we will substitute dysfunction with negative X instead. So rewriting this limit as experts needed to two minus native X or plus X divided by two plus X Now notice that I didn't write experts in ecstasy from the left or expert unit two from the right because both of those give you the same results right since we're pushing native to which is a value in this region. So the function looks the same, whether it's from the left or the right. As we can see, we can cancel these too, and this simplifies to one And if we take the Limited's experts think, too, of this constant value of one. Her answer for the limit is one again keeping in mind that this is true for both. Pushing is equal to negative to pushing that from the left and from the right. And since those Walter True and exist and they both equal one, then this limit from the beginning also equals one, and that is our final answer.

#### Topics

Limits

Derivatives

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp