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22y6 Show that lim does not exist (2,y)-(0,0) 524 + 2yl2...

Question

22y6 Show that lim does not exist (2,y)-(0,0) 524 + 2yl2

22y6 Show that lim does not exist (2,y)-(0,0) 524 + 2yl2



Answers

Show that $$\lim _{(x, y) \rightarrow(0,0)} \frac{x y+y^{3}}{x^{2}+y^{2}}$$ does not exist.

Well we have a question and it is We need to find the limit or we have to show that it does not exist limit X approaches to minus infinity. Okay, access squad Place to express to the Power seven. This could Burton is limit X approaches to minus and pretty Let us take access to about seven is common. So when by express to the par five plus two. This could also be written as limit x approaches to minus and 30 Accident. The power seven plus limit X approaches to minus infinity one by accident about fine blessed two. No, we see that uh if this is always equal to OK, let us right limit X approaches to minus and 50 X raised to the power seven. Bless too because this will When explosives to my sincerity this whole thing will get close to -02 or zero. Okay, no, since this is Exodus with Power seven which is odd power. If this has odd powers So if X will approach two miles and 20 X raised to the power seven, we'll have to close to minus infinity. Okay, so everything Will be will oppose to minus and 20 minus infinity. They should be the answers. Thank you

To evaluate this limit me 1st factor out the variable with the highest exponent for the numerator and variable with the highest exponent for the denominator. That means we have limit as X approaches infinity of in this case the numerator well have X cubed as its variable with the highest exponent. And so we get X cubed times. We have four plus six over X two over X cube. And then this will be divided by um for the nominator. We also have X cube as the variable with the highest exponents. So we have X cubed times two minus, pour over X squared plus five over X to the 3rd power. Now simplifying this, we can cancel out the X cube and we have the limit as X approaches infinity of four plus six over x -2 over X. Cube. That's divided by 2 -4, x squared Plus five over x cubed. Now evaluating at infinity we have four plus six over infinity -2 over infinity over 2 -4 over infinity plus five over infinity. Now Know that a constant over infinity will always approach zero. And so six over infinity will approach zero as well as to over infinity for over infinity and five over infinity. And so we are left with 4/2 which reduces to two. And so this is the value of the limits

To find this limit. We first want to rationalize this. So we multiply this by the conjugate of this one. That would be the square it of four X squared plus three x minus two X. This all over the same expression square it of four X squared plus three x minus two X. And so from here we have the limit as X approaches negative infinity of we have the square of the square it or four X squared plus three X minus the square of two X. This all over square it of four X squared plus three x -2X. Now simplifying, we have the limit as X approaches negative infinity of four X squared plus three X minus for X squared this all over square it of four X squared plus three x minus two X. I'm simplifying further, we have the limit as X approaches in negative infinity of three x. Over we have square it of four X squared plus three x -2 x. and then and here we factor out the variable with the highest exponent for the denominator. And we have limit as X approaches negative infinity of three X. This all over before the square root we have squared of X squared times four plus three over x -2X. And this gives us limits as X approaches negative infinity of three x over negative X times the square of four plus three X. Now this is negative because our excess approaching negative infinity. So square root of x square This is the same as negative X squared and taking the square it off. So this should be negative X. And then from here We have -2 x. And then we factor out negative X in the denominator. So we have the limit as X approaches negative infinity of three X over you have negative X times square it of four plus three over x minus negative two. Or there will be plus two. And so we can cancel out the X. And we have the limit as X approaches negative infinity of negative three over square at the four plus three X plus two. And here, yeah, evaluate as infinity and we get negative three over Square it of 4-plus 3 over infinity plus to this will give us negative three over the square root of four plus zero plus two, which will give us negative three over two plus two or negative 3/4. And so this is the value of the limits.

Prove that limit. X com A light and into 00 off three years Girl rode off. Why do I didn't buy extra? The problem before Bless by a squared does not exist. Gotta stick along X is equal to zero We get limit zero comma by telling zero Bama 03 year zero square root Why divided by the road to the powerful plus by scare equals You don't buy my squid Equis zero Now we will check along by the girl Do X squared So we get them it Expo my X squared and in tow zero Parma zero three X square into route off expert divided by X to the power of oil plus export to the power square The supplies t A X squared into X divided by do X to the bubble. For we get three by to therefore limit. Do not agree, which implies the MIT does not exist


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