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Question 3- Prove that sin R = R is uniformly continuous. [You may assume without proof that Isin(z)| < Ixl and sin(x) sin(2) = 2 cos(%+2) sin(834) for all T,2 ...

Question

Question 3- Prove that sin R = R is uniformly continuous. [You may assume without proof that Isin(z)| < Ixl and sin(x) sin(2) = 2 cos(%+2) sin(834) for all T,2 € R] 6) Prove that r(z) = 1/z is not uniformly continuous on (0,00) , but is uniformly continuous on [1,0).

Question 3- Prove that sin R = R is uniformly continuous. [You may assume without proof that Isin(z)| < Ixl and sin(x) sin(2) = 2 cos(%+2) sin(834) for all T,2 € R] 6) Prove that r(z) = 1/z is not uniformly continuous on (0,00) , but is uniformly continuous on [1,0).



Answers

In Exercises $7-16,$ use the Laws of Continuity and Theorems 2 and 3 to show that the function is continuous.
$$f(x)=x+\sin x$$

Proof is to ICO and use that we value is the problem. So here we first observed that there's a should not Dante Cose I x is a sem aside pie over to minus X. That means thiss should be f off side pi minus two over minus X. Here we use u substitution. You know, most pie over to minus Next you is most active. Thanks. Way open another page and see what it gives us. So U equals pi over to minus X So excess, Cyril, you is you Is pi over too? When excess pirate too use you and this F off sign of pie over to minus X becomes f off. Sign you and the ex It is the CMAs negative to you. Sorry. Here, here. We know if we swapped after bond or bomb. Which way Beauty Private. Another negative sign. So this because to this a U N ex actress name of variables like right this So we start with we'LL start with the formula on the left hand side and and we got the formula on the right hand side. So we prove the identity which is the first part They use the first part to evaluate this integral oil On the other page here it looks like if we pick x square f of X equals x square, then we'LL have It's a girl from zero to pie over too Sorry, we used part and pig If off because x squared then in this case it keeps us by the conclusion ofthe party We know that this and this it's a simple you look at the formula here f f it's x squared that eleven size coz I square the right hand side sine squared So sorry, here's not pie should be pi over too And that said this too high Is there a easy way to compute the eye? Well, we know coz I square is is one minus sai square So this also echoes zero to pi over two one minus zero two pi over too size square Next e X which far definitional Hi! And this are supported the Grover Constant because this is a pie over two minus I So here we got the equation. You know, we get the equation. Hi, because time over, it's you minus side, which means I course two eyeholes pyre or two. So I host Pi over four and that is our final interviews with Defi i Toby Deceit Group

Okay, so the function we have here a part A is F of X is equal to sine of X plus Y over X plus Y. Now we know the function F is continuous at X is equal to A If and only if we have the limit as X approaches A. Of F of X is equal to F. Of A. So here we observe that as X Y approaches 00 the expression X plus Y must approach to zero. So let's let X plus Y be equally tea. So let X plus Y the equal to T. And then we have well um clearly we then had the T approaches zero. So we have the limit as X Y um approaches 00 just becomes the limit as T approaches zero of instead of sine of X. Y over X plus Y, we just have signs of T over T. And we know the limit here is equal to one. So therefore the original function um this is going to be equal to one And therefore the function is continuous at 00. By defining f of 0, 0, be equal to one. Okay? And then for part B we consider the expression F of X, Y is equal to x, Y divided by x squared plus Y squared. Ok, so um here we find the limits of F along first the path Y is equal to uh M X. So we have here the limit as X Y approaches 00 of x times M X over X squared plus M X squared, gives us just the limit as X approaches zero of em X squared over X squared plus um square times X squared, which is just equal to M over one plus M squared. And since here M varies in different paths. The limit values. The function are different in different paths. For example, along the path, Y is equal to x. Um We get the limit is equal to one half and along the path Y is equal to zero. Um We get the limits equal to zero. Therefore the limit as X Y approaches 00 of F. Of x, Y does not exist. And um last lethal part C. We have the function F mapping are square to our where we have X. Y maps to Y times E. To the X plus sine of X plus X. Y to the fourth. Okay so the function here is some of three functions and now it's gonna be sufficient to show that each term is confession is um is continuous. Now since we have at the limit as X Y approaches, sign of acts is eager to sign of A. Um And also X. Y to the fourth is a polynomial of two variables is continuous at all points. Um Therefore our function here is going to be continuous.

Today we will be using the laws of continuity. And there's two and three to prove that the falling function why equals X minus X is continuous. So well, I will do first is I will break this down into the two serums that I mentioned before so X can be considered a polynomial because it is the same as, um, why people's tax to the one. And when you have a power of a non negative tischer, it can be classified as polynomial. And we know that for this given polynomial, the domain is all real numbers and there is a given why for each given acts. So this has to be continuous. Um, and given that Sine X is a basic function as well, it's a trick function. According to the urn three, it's continuous on its domain, which is also all real numbers. And so the fact that the product, uh, these two basic functions which are continuous, um, means that this whole function within itself is again continuous, which addresses this basic law continuity. So that's how we know that this function is continuous based on all these three components

All right. The problem is asking us to use the laws of continuity and theorems 10 3 to show that this functions could take us, uh, dysfunction being X plus sign fix. So serum three tells us that why equals X to the one over end for some and in the natural numbers is continuous. Now, notice that one is a natural numbers on natural numbers air the positive integers whole numbers so X to the one over one is just ext one, which is X and one over one. That one is still a natural number. Sorry, that's a lot of ones, but that's what it ends up working out, too. And staring three also tells us that why equal sign next is continuous. So just from three and three, we know the two parts of this function our continuous. But now we just need to check whether there some is continuous and that's where the basic laws of continuity come in. And basically, what that says is that if the two parts of a function are continuous, then the sum is continuous. So if X is continuous, which had his best year in three and signed X is continuous than X plus sign acts which is equal to ever vex is continuous and that's it


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