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In Exercises 2.4.2-2.4.40, find the indicated limits:...

Question

In Exercises 2.4.2-2.4.40, find the indicated limits:

In Exercises 2.4.2-2.4.40, find the indicated limits:



Answers

In Exercises 21–28, find the limits by substitution.
$$
\lim _{x \rightarrow 0} 2 x
$$

In this problem. We want to get the funny off the limit off to eggs. When X approaches to boys substitution, we can get to buoy two equals four. And this is the answer off this problem.

Okay, so we have the limit here as X approaches zero of Tangent of X. Well, we know the tangent of acts is just sign of acts over co sign of acts, right. We could rewrite this as the limit as X approaches zero. While tangent is just sign over co. Since we have sign of X over co sign of acts. Well, if we just, um, plug in zero here to take the limit, what do we get? We get well, the limit off sign of zero, which is zero over co sign of zeros. We get 0/1, which is just zero. So, um, therefore, the limit as X approaches zero of tangent of acts is just equal to zero zero.

The given problem, we want to estimate the limits and the exercises graphically. So we know that the limit is going to be as X approaches zero of the absolute value of X over X. With this in mind we consider the fact that if we plug in a zero here and here, it ends up becoming undefined. Um and in fact if we consider the absolute value the graph itself, we have the value of X divided by X. What we see is that We end up just getting a straight line of one um up here and then we get a straight line of negative one down here. But we see that based on this, the limit as we approach from the right is a positive one, but the limit as we approach from the left is a negative one. And because of that, we see that the limit cannot exist. So we see that it does not exist.

Hello, guys. In this problem, we have the limit of secret X as X approaches. Zero think No, this is Secret. X is equivalent to won over call sign X. So we're gonna self zero with how one over cosine zero. And they note because I kn zero is equal to one or one over one. We have the limit equal to one.


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