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HWO4 11.1-11.3: Problem 6Previous ProblemProblem ListNext Problempoint) In this problem we show that the functionIxl _y f(xY) = x2 + ydoos not have Ilmit as (x,Y)(0...

Question

HWO4 11.1-11.3: Problem 6Previous ProblemProblem ListNext Problempoint) In this problem we show that the functionIxl _y f(xY) = x2 + ydoos not have Ilmit as (x,Y)(0.0)(a) Suppose that we consider (x, Tx? _ y lim (521')-(Q,0) +>along the curve2x" . Find the Iimit in this case:(b) Now consider (x,Y)(0,0) along tha curve3x2 Find Ihe Iimit in this case:lim14_ (1,Jr-)-+(0.0) x + y(c) Note that the results from (a) and (b) indicate that has no limit as (X,Y)(0,0) (be sure You can explain

HWO4 11.1-11.3: Problem 6 Previous Problem Problem List Next Problem point) In this problem we show that the function Ixl _y f(xY) = x2 + y doos not have Ilmit as (x,Y) (0.0) (a) Suppose that we consider (x, Tx? _ y lim (521')-(Q,0) +> along the curve 2x" . Find the Iimit in this case: (b) Now consider (x,Y) (0,0) along tha curve 3x2 Find Ihe Iimit in this case: lim 14_ (1,Jr-)-+(0.0) x + y (c) Note that the results from (a) and (b) indicate that has no limit as (X,Y) (0,0) (be sure You can explain whyl): To show this more generally; consider (x, Y) (0,0) along the curve mx" , for arbitrary m . Find the Ilmit In this caso: Jxl _ y lim (I,m" )-+(0,0) x2 +y (Be sure that you can oxplain how this result also Indlcatos that has nO limit as (,P) (0,0).



Answers

Show that the limit does not exist by considering the limits as $(x, y) \rightarrow(0,0)$ along the coordinate axes. $$ \text { (a) } \lim _{(x, y) \rightarrow(0,0)} \frac{x-y}{x^{2}+y^{2}} \quad \text { (b) } \lim _{(x, y) \rightarrow(0,0)} \frac{\cos x y}{x^{2}+y^{2}} $$

Problem. Seven. Here you and function is limit. It's come out by 10 to 0 comma zero, which is a record of three by X squared plus two ways quit. So if you do well, it's physical. Zero. It would have been indignant Why it tends to zero. Do you different? Zero square us to my eyes. So which will be limit like That's 20 What do you buy below vice. Quit with musicals. No. Do fight thoroughly. It does not exist. Let me That's no exist and that the Beeb artist limit exclaim, Abi tends to zero comma. Zittel is the express. Why developed by two X squared plus all right square. So it will be like living. I know ecstasy. Korda's it'll. We have congregated. Why tends to zero X place like verbally? Do we do their squared plus y squared, which is a cultural limit? Why it tends to zero what by visiting cargo mark do fight so limit does not exist

In this problem of limit and continue to we have to show that the limit does not exist. So we have given that limit two X square plus why is square divided with excess. Grad plus white squares provided that ordered pair X and Y approaches towards the 00. And now first we will find the value of the limits along the line. Why is equal to X. So for this we have to put a Y equals two X. So here this will it would be now putting ways equals two X. So this is X approaching towards zero. Now putting to access square this where they would be plus X squared divided with X squared plus X squared. And now when we solve it this is equals two three X square limit extents to zero. This way there will be three X squared divided with two weeks is square which is equals two from here excess where taxes were cancelled out and the value of limited three D. Would be too. And now we will find the value of the limit along the line. Bicycle tour attacks. So this age limit here X approaching towards zero. Now this where there would be two X square plus we have to put Y equals Bluetec. So that's where they were We do takes holy square which is X divided with X square plus X. And now when we take ex comin from the numerator and denominator we would have a limit X approaching towards zero and X. This way there will be two X plus one divided with ex common and we have X plus one from here. Extra X can be common, cancel out. And now when you put X is equal to zero. So we have the video equals to one. Since the value obtained on the line, Y equals two X. At the point X. And Y equals to 00 is not equal to the value obtained on the line. Y equals the root tex. When we put x, Y equals 00. So that's why hence we said that. Hence we say that limit does not exist. Limit does not exist because both are two different value that is here three widely to and this gives one. So that's why we see their hands limit does not exist.

This problem wants you to find the limit of X approaching zero for seven minus two squared a four minus X squared all over X. And in my little table there it is using some scientific notation. But those numbers are getting closer and closer to zero. They're getting very very very small as it gets closer to zero. And I can trace that and see that on my graph here. So from the left, as I'm getting closer to zero on the X, it is getting closer to zero on the wire output value. And from the right same thing getting closer and closer to zero. So we'll see if we can verify that algebraic lee. So I'm gonna start by rationalizing the numerator. So this will will multiply by seven plus the square root of 49 minus X squared. Um We're just multiplying by one. So I have to multiply the top and bottom by the same here. So on the top, when you do foil you'll get seven times seven is 49 plus seven times Route 49 minus X. Where minus seven times Route 49 minus X squared, which gives you zero of those. So then you're left with minus the square root of 49 minus X squared times the square to 49 minus X squared, which is 49 minus X squared. Simplify that. You get 49 -49 um is zero and minus a negative X squared. Is positive X squared. For the top. On the bottom, you'll be left with X times this group of seven plus the square root of 49 minus X squared. Now, what can I do on the top? I can cancel out an X from the X squared. Put the X on the bottom and that limit has no restrictions on its all have X going to zero as X approaches zero have X over seven plus the square root of 49 minus X squared. And whenever there's an X. Or wherever there's an X. I'm gonna pop into zero here, so I'll get zero. And on the bottom here we'll get it zero. So as long as the bottom doesn't equal zero, then we'll be good. So we get zero on the top over, it will be seven plus Route 49 which is seven, so 77 is 14 were divided by 14 gives us a limit of zero.

This problem wants you to determine the limit as X approaches zero for the square root of seven plus two X minus the square to seven all over X. So it appears according to my table that we're going to get some decimal number around 0.37 seven or 378. So and let's verify this on the graph. As I'm resuming in here, I can trace says approaching zero. I do get closer and closer to that. 0.377 or zero 378 from the left and from the right. So let's see if we can calculate this algebraic lee. So again, I'm going to start by rationalizing the numerator so I'll multiply this by the square of seven plus two X plus route seven. What I do to the top while I'm going to do the bottom because I'm just multiplying this by a weird version of one right? So foil the top, multiply these two together, you'll get seven plus two X. Then you get Plus Route seven times route seven plus two X minus route seven times Roots and Plus two X. Which is just zero of those. And then you have negative route seven times route seven which is seven which leaves us on the top with just two X. And on the bottom. I'm not even going to distribute that. You'll just have X times this group of Route seven plus two X plus route seven. And that means I can cancel out these exes. So I'm left with the limit As X approaches zero of 2 over The square root of seven plus 2 X plus Route seven. Now let's substitute in that X as a zero here making this The limit of two over the route seven plus route seven because two times zero is just zero. So what I did here is I just made this two times zero for ex. And what we get then is a decimal answer. If you use a calculator for that of 0.37796, which verifies our graph.


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