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QUESTIONCompute the following limit, Or show that it doesn't exist_ cy lim(z,y)-(,0z2+y...

Question

QUESTIONCompute the following limit, Or show that it doesn't exist_ cy lim(z,y)-(,0z2+y

QUESTION Compute the following limit, Or show that it doesn't exist_ cy lim(z,y)-(,0z2+y



Answers

Find the limit or show that it does not exist.

$ \displaystyle \lim_{x \to \infty} (e^{-2x}\cos x) $

Suppose you want to evaluate this limit. I know that if our function is a rational function involving E and our limit is that infinity the first thing you have to do is to factor out E was the highest exponents. And so from here we have limits as X approaches infinity of in this case between E raised to three X. And erase negative three X erase the three. Access the one with the highest exponents. So we have erase the three X. This times one minus, erase the negative six X. And then this all over the same process. We have erase the three X times one plus He raised the -6 X. From here we can reduce by getting rid of the common factor, erase the three X. And we have the limit as X approaches infinity of one minus erase negative six X Over one plus erase to -6 X. From here we Evaluate that infinity we have one erase to negative six times infinity over one plus erase to -6 times infinity And this is equal to 1 -1 erase the negative infinity Over one plus erase the negative infinity. And we know that erased the negative infinity approaches zero and so from here we have 1/1 which is equal to one. Therefore the value of the limit is one

All right, we want to find this limit or show that it doesn't exist. Um, a lot of people think about using local tiles rule or thinking about and behavior of functions to solve these. Um, I don't like those. I think Loki tells rule, well limits have to exist before derivatives. So this is a more fundamental question that and then end behavior can be hand wavy and or confusing. So um, these are both growing, this is really big, probably really big negatively. This one's really big. Um, so let's just stop them from growing. That's going to be the least confusing. So I'm going to divide top and bottom by E to the X. So this is one over E to the X minus italy X divided by either the X one and then here we have one over either the X plus two, you do the X divided by either the access to and now this theorem I'm going to write down is one over eat the X is going to go to zero. Why? Because X is really big. Even the X is really big. Let's just have this sketch of the X in the back of our minds. You know, the X is even bigger and then one divided by really, really huge number as the number gets huger is going to go toward zero. So now we can just use that there we have the top That goes to 0 -1. So the top goes to -1. The bottom goes to 0-plus two is to, So we get a limit of negative 1/2. I think that's the most straightforward way of doing this kind of problem.

All right. Um We've got a really cool limit question. We've got our tan of either the X. That's fun. Um Okay so we need to think about what tan and our town looked like. So let's just look at tan first tan of zero is zero And then between negative pi over two and pi over two we get up to and actually it doesn't go flat. The difference between tan and X cubed is whether they go flat at the charge, there's more differences but for sketching purposes. Okay, so this is um Tan X. And there's a there's a lot more of them. But we don't care because when we take the inverse function we only take one part of the domain so that the inverse is still a function. Otherwise once we took the inverse that wouldn't satisfy the vertical line test. So now we're just going to switch the X and Y axes or reflect across the line by equals X. Whichever way you prefer to think of it. But we're only gonna do this piece and then this is this is the function. Um Oh, sorry about that. Why equals arc tan X. Okay. Now, um what happens as X? It's really big? Italy excess is even bigger, right? Just think about have this graph of E to the X in the back of our minds to this is why he calls me into the X. It's very, very big. Um Okay, so we're looking for are tan of a big value of X. Arc. Tanne of a big value of X looks like that's going to be pretty close to pi over two. So, um we just solved the season grass so that equals pi over two. Yeah.

So here we are given information about specific limit. We have the limit as X approaches infinity of one minus X squared over X cubed minus x plus one. Yeah. So mainly here, once again, since we're evaluating X as X approaches large numbers, the constant terms are relatively going to be insignificant. And also when we compare X cubed and X X cubed is going to be significantly larger than X. So we can ignore X as well. So in the end this would just be evaluating the limit of the form negative X squared divided by X cubed and we can cancel out X squared from the numerator and denominator. And we just end up with negative one divided by X. So this is equivalent to negative one divided by infinity. So one over the within these just zero. So this would be our answer. However, we can also evaluate this by lawful rule which involves differentiating the numerator and denominator. However, this would just require more iteration. So it'll be -2 X. This will be three X squared minus one. Then we would have to do another iteration of law of the rule where we can find we would have negative too and this will just be six X. And once again we can apply Direct substitution and this would be negative to over six times infinity Which would just be equivalent to zero and this is our final answer


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