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& 1 Prove lim exists and equals zerom (u)h (oou 22 +y2...

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& 1 Prove lim exists and equals zerom (u)h (oou 22 +y2

& 1 Prove lim exists and equals zerom (u)h (oou 22 +y2



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Prove rigorously that lim sin $\frac{1}{x}$ does not exist.

Mhm. We can still choose Georgia echoes a minimum of one and absence in the hands. If The absolute value of X zero is less than Data and creators and zero then we can imply the absurd of the value of extra force times signing X squared one over X minus their role. Echolls the episode of tomato of X times the absolute value of X cubed times the absolute value of science squared while racks and those that this term is less than third to This term. Is that so equal to one. And this term is also that's an equal to one and hands. Yes. Yes. Nason sorry. This expression is less than dirt to and since go to is less than or equal to answer No. Yes yes yes now equal to Williamson. So this means the limit as X approaches their role Of this expression equal zero.

In problem. Six. We want to evaluate the limit over function off two variables when we approach one and minus one. We can find this limit by getting any value over the function. When we approach this point from any direction and we can use direct substitution because we have here to multiplied functions, the exponential function and the cozy in function which they are both continuous everywhere. This means the resulting function is continuous everywhere. Then we can use direct substitution off the values of X and y. Then it equals he would to the bar off minus one multiplied by minus one. Multiply by co sign X plus Why we have one minus one equals e to the power of one multiplied boy co sign zero, which is one co sign zero, which is one this makes the value equals e to the bottom one, which is e on. We can right the value off E, which approximately equals 2.7 one. It

We were asked, Approved the limit as X approaches zero on sign of one of Rex does not exist using the formal definition of a limit the non existence of something. It's usually a good idea to so, using proof by contradiction. So suppose the limit Exit purchase zero. One sign of one of Rex is equal l for some rial number. No. Now let's try to visualize this craft so we can figure out, um, what the gap size is going to be for a given epsilon and then try to take a suitable Absalon says that we confined an ex for any adulthood such that that X is going toe. The FX will lie outside the gap. So, um, outside the absolute neighborhood. So the graph sign of one of her ex. So if we think about the limit is X approaches Infinity unsigned of one of her ex. This is going to be assuming parliament exists. Sign of the limit is expertise infinity of one over X, which wouldn't be sign of zero, which is just one so or zero. So we expected as we go to infinity sign of one of Rex should approved zero and Likewise, as X approaches Negative infinity. And so we have that exes perching, affinity. One of her exes approaching zero from the right. So we have it for sloping down towards infinity. And then if X is less than zero, then okay, a limit is expertise. Negative. Infinity of one of her ex is going to be approaching zero from left. So from negative numbers, Uh, and in between, we have something that looks kind of like a sine function at first, but rapidly, it's tighter and tighter and tighter. Delegates to the origin. I don't know if it's gonna be graphic. A little bit messy, mimic its less tight until eventually we get back to more of a, uh, open Kurt. This is our function. Sign of one of her ex approximately. And so let's consider different values of l. So clearly we had that. If l is say, here we can pick up salon to be nips Alaniz say, um, this large so that for underneath the function, then we have an issue because then it's pretty clear that for every delta green and zero, all the X um satisfy ffx minus l will be less than Epsilon So what if we try a different absolute? So if something needs to be smaller than that because that Epsilon was just big enough to reach the bottom of this curve, that would mean that the bottom of this curved, by the way, is when sign of one of her exiting via its lowest, which is negative one. So in this case, we were the Epsilon would be the distance between negative one and L. In this case, he said, there would be infinitely many points on the bottom of this function sign of one of Rex such that f of X would not be in an excellent neighborhood of l. And so we see that if we take any Delta neighborhood of zero will be able to find one of those points. So in particular, this is going to apply as long as l is going to be greater than negative one. So for one case, let's suppose that l is greater than negative one and we want to pick EP Salon to be equal to at most it should be. The difference between L and negative one is going to be absolute value of L plus one which we see from our inequality is going to be graded zero since is simply going to be helpless one. So we want to find Delta such that sighs of the gap for some X in that Delta neighborhood is going to be greater than or equal the absolute. So I guess, let's investigate the size of the gap we have that this gap is going to be sign of one over x minus l and we're looking for an ex such that this is going to be greater than or equal to. It's a one. Okay, so this means that sign of one of Rex should lie between No, we're not between. That should be greater than or equal to l plus Epsilon or sign of one of Rex should be less than or equal to l minus. Absalom. Now we have that because l is greater than negative one. We have that and Absalon is l plus one. This is going to be l plus one plus l is to l plus one and l minus. Helpless one is going to be negative one. So it's not clear whether 12 plus one is going to be less than or equal to one. But it is clear that sign of one of her ex is going to be less than or equal to negative one. Because if sign of one of her ex is going to be less than or equal to negative one, we have that sign of X. One of her ex can be at the very least, negative one. So the supplies sign of one of Rex has to equal negative one. And so we have that one of her ex is then going to equal to the inverse sign of negative one. We know there are infinitely many points like this, and they have the form pi over two or negative pyre or two plus two and pie where an is an element of the intruders. And now the question becomes, Well, how do we pick a delta so that we can find an X inside the delta? Well, we're not picking it all to actually, so just Delta is just any really number girded zero, and we want one over X, the absolute value of this or with the absolute value of X, you between zero and delta. Well, we know that this implies access to lie between positive and negative Delta. And this tells us that since positive negative Delta are both non zero one over X, it's going to be huh? Instead of seeing it this way, maybe we can solve this or X that X is going to be equal to, and this is going to be to end. Times two is four n minus one over to for end minus one over two pi for one over X. This implies that X is going to be equal to to over four n minus one. Hi, where n is some integer And so this inequality then tells us that, yeah, that zero is going to be less than the absolute value of two over for end minus one. Hi. Listen, Delta complies. A zero is less than two over the absolute value of four n minus. One times pi people less than Delta. So I think figure out what values of n work one is for end minus one going to be greater than zero. This is one and is greater than 1/4 so and must be greater than or equal to one since it's an intruder. And so we have that ends great than or equal one zero be less than to over four n minus one times. Pi will then be less than Delta. And so we have that. Yeah, these two inequalities tell us that left hand side, nothing On the right hand side we have that or end minus one must be greater than two over. Hi Delta. This implies that and must be greater than two over Pi Delta, plus one times 1/4 is equal to two divided by four. That's you want over two pi Delta plus 1/4 and we have that. This means that and is going to be the integer must be greater than or equal to the inter. That's just above this, which is going to be 1/2 pi yell toe. So as long as we choose any end, that's greater than or equal to one over two pi Delta should be OK. It has to be greater than that, and it has to be greater than or equal to one. So we'll choose end to be equal to. Let's see, I wanted to be the maximum, uh, one and 1/2 pi Delta. You want to say that I actually mean me, Ceiling. Uh, we're going back ends greater than that. It has to be greater than or equal to 1/2 pi Delta plus one. So I want end to be the max, uh, one and the ceiling of 1/2 pi Delta plus one. So what does this all mean? Let's do a formal proof for if l is greater than negative one. Then we have that we're going to pick up salon equal to we said Epsilon should be l plus one just clearly going to be greater than zero and then let Delta be some positive number. And then we're going to pick in, which is the max, uh, one and ceiling of 1/2 pi Delta plus one and number going to pick X to be equal to to over four and minus one pie. Then, as we demonstrated above in calculating in and X, we have that, huh? We have that zero is when the lie between X and Delta so but we have that the size of the gap f of X minus. L is going to be Well, this is gonna be sign of one over X in this case, Waas going to be four n minus one over two times pi minus l and we chose en recall so that this was going to be true. And then we have that for this value of n read. More clearly, this is the sign of gonna be two n pi minus pi over two, which is equal to negative one. So my new cell is equal to opposite of l plus one. Yeah, the l plus one is greater than zero because she just right, this is opposite of F salon, just simply equal to Absalon. So we've shown that are gap actually has a size of at least absolute so it follows that the limit X approaches zero. Well, I'll say this is a contradiction. It's And now well, suppose that the opposite is true instead of l being greater than negative one. Suppose l is less than or equal to negative one. So looking back at her picture, this means that now the blue line it's going to be below the function and we see that they're different values of epsilon we can take so you could take up salon to be smaller than this is between l hands negative one. Unless L is equal to negative one between included as a possible case. So in that case wouldn't really be able to use that EPPS warm. So let's think a little more generally well, suppose that Absalon was just big enough So that Absalon bridge to distance between white gazelle and why equals one? Then we have all these points on top of the curve of the end, the Epsilon neighborhood, or would be on the boundary of the Epsilon neighborhood of why and so the limit would not exist. So in order to formalize this, let's take Absalon. Be equal to this will be the distance between El and one. We have all minus one is less than a gram Negative twos. This is going to be equal to the opposite of L minus one. Just negative, ill less one. This is clearly going to be greater than zero. Now, how are we going to pick the delta that we want? Well, as in the previous part, will have to find next value such that sign in one of her X equals this time instead of negative one. If you want sign of one of her ex to be equal to positive one. So this implies that whenever X is now equal to positive pi over two plus two n pi, which then complies, it's is going to be four end plus 1/2 pi. And that means that X is going equal to two over foreign plus one pie. And because we want extorted between zero and Delta, this implies that two over foreign plus warm pie has been less than delta. So we have that when a fan quote that you have ends foreign plus one grated zero. Well, this is when n is greater than negative. 1/4 in the end must be greater than or equal to zero instead of one. So now we have and is going to be greater than or equal to zero. In fact, occurred to me that in the previous part and should be Max Okay, good. So we have that as long as and is greater than or equal to zero. Yes, that zero was less than and over two over a foreign plus one pie just less than Delta means that foreign plus one is gonna be greater than two over Pi Delta, which means that N is going to be greater than 1/4 two of a Pi Delta minus one. She is equal to 1/2 Pi Delta minus the fourth. And so we want to pick end to be greater than or equal to. I think I'll keep in like this the ceiling of one of her two Pi Delta minus 1/4. And so this time we want to take en or before we take and let's let don't Toby some positive number and let's pick in to be equal to the max um zero and the ceiling of 1/2 Pi Delta minus fourth. Okay, and we'll pick X to be equal to to over four n plus one pie. Then it follows from our calculations that the absolute value of X lies between zero and Delta, but size of the gap sign of one of Rex minus l is equal to sign of. This is going to be four n plus one over two pi minus l, with the sign of foreign plus 1/2 pi since and is an integer, this is just going to be one minus l and recall that El is going to be when we picked Epsilon to be one minus l. So this is equal to the absolute value of F salon? Absolutely. Of course, greater than zero suspend equal to Absolutely So you've shown that besides of the gap is exactly upside and therefore we have another contradiction. But we exhausted all possibilities for l to be chosen all l above and below negative one, including native one. This implies with limit as X approaches zero on sign of one of Rex does not exist And this is what we wanted to prove.

And so we can adjust this question by considering the linen, uh, along two pervs approaching the value of X Y equal to one and one s O for one value, we can consider approaching on the curve. The hyper villa next is one of mine, and in this case will have one of the white touched my squared. Plus one for one, huh? In the course, let's quit over. Why is lying wireless one over five? Unless one over. Why last one? Which one? Now? Um, another curve we can consider is approach along the vertical line X equal to one. What this implies is just one time. Do I swear? That's all, uh, again. Well, while this one on, then we could observe that one squared plus one. Because while A s are in terms of my plus one, we can cancel the two instances of my last one. Yeah, and then finally calling the why Swells X is converging to one. We're left with value one plus one, just two. And, uh, this shows us that the one that does not exist one, because when it does, it has to exist in apparently the path along When shit's approach on. Yet here we found two pats on over The proble X is one over why one of the brutal Klein X is one in which case the candidate limit values respectively, are one and two wishes are equal, which are not equal. Um, until that implies again, it wanted that the limit does not exist.


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