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Prove or disproof the existence of the limit:lim...

Question

Prove or disproof the existence of the limit:lim

Prove or disproof the existence of the limit: lim



Answers

Find the limit or show that it does not exist.

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

Trip with poor the sex. Yeah, because here for one day, you the super she's gonna be given that b and grand here and tell that to be creative with the euro, we have zero. He's ex remains zero capsule. This is Delta. Wonderful. Absolute X B. So this is here. So absolute x Boston won t explain. Cereal was less than on predatory. Be so that you moved that X delta implies that one by one of our like excess better than deep. So this cruise, then the 01 Teoh

Your limit. Experts issue for Wonderland X square perches Nega baby, We can improve that first you have low who via X minus zero is held idea here and with the statement made at one point, we just list them name. This is the given first we rearranged. So I think one of the X squared is craving Be expert ex weird thing one of Avery X is supposed to be won by someone beat. Absolutely wonderful basic question would be that Delta is going hard square. Group E a negative X squared is less than negative so that it implies that the animals forget negative He has beaten last and zero was the show's perp cruise through and excessive your one x squared equals.

We were asked to prove that the limit is expert zero in absolute value of X equals zero. Using the formal definition of a limit. Do this. Let's consider the size of the gap. This is going to the absolute value of absolute value of X minus zero, which is just absolute value of X. And, of course, the absolute value of nervous. The value is just the absolute value. So an absolute value of X and we know that absolutely of X is going to lie between zero in Delta, this would be less than Delta. We want to take a delta such A This is less than or equal to absolutely well that me, one delta to be any number between zero and excellent. So for the formal proof. But Absalon, be some positive number must take Delta to be apps on. Clearly dealt is greater than zero. And we have that if absolute value of X lies between zero and delta, then as above, we have size of the gap is equal to the absolute value of X just less than delta, which we have is equal to absolutely. Therefore, you have shown that the size of the gap is strictly less than absolute. So that limit as X approaches zero on the absolute value of X is indeed equal to zero. This is exactly what we wanted to prove.

We need to prove that the limit as X approaches zero of sign of one over X does not exist. So let's let Delta be greater than zero and l be in the rial numbers. So to prove that the limit does not exist, let's prove that. So this isn't a side. This isn't part of the proof. We will prove that this sign of one over X minus l is greater than or equal to one half for some X such that X is less than delta. So this is using the absolute in Delta definition of the limit. Okay, so sign is an oscillate torrey function. So it also it's between one and negative one. So we need to deal with that. So what? And be greater than zero be a and is in the positive integers. Okay. And we need in large enough such that two over four in plus one pie is less than delta. So why four in plus one pie? So you'll see that if we have sign of one over X one, ID equals two. Sign foreign plus one pie over to than in equals one. And if sign at one over X two is equal to sign of foreign plus three pie over to this is a negative one, so we found two bounds. Now, if the absolute value of sign of one over X one minus l is greater than or equal to one half, then the limit does not exist. If sign of one over X minus. L is less than one half, then we have negative. One half is less than sign of one over x minus. L is less than one half and add elta. Both sides al minus one half is less than sign of one over X is less than l plus one half, and we have that sign of one over. X one is equal toe one. So we have that one is less than l plus one half. So that means l is always greater than one half. And we have the absolute value of sign of one over x two minus. L is equal to the absolute very negative one minus L, which is greater than three halfs. Either way, their existing X such that the absolute value of X is less than delta. But the absolute value of sign of one over X minus l is greater than or equal to one half, so no limit l can't exist


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