Find $\lim\limits_{(x,y) \to(0,0)} \frac{xy^2}{ x^2 + y^4} $

Question

I am doing some exercises for my Calculus 3 exam and I get stuck in this exercise: (I need to find the limit or say that it does not exist)

$$\lim_{(x,y) \to(0,0)} \frac{xy^2}{ x^2 + y^4} $$

I tried to change it polar coordinates (so that I have only one variable) but it is messy and I dont understand it...

Can someone explain me how to do it? Thanks!

http://math.stackexchange.com/questions/174190/prove-that-lim-limits-x-y-to-0-0-fracxy2x2-y4-0?rq=1 — Amzoti, Mar 30 '13 at 00:21

score 9 · Accepted Answer · edited Apr 13 '17 at 12:20

9

Approach $(0, 0)$ along $(t, t)$ as $t \to 0$ to get: $$ \lim_{t \to 0} \frac{t^3}{t^2 + t^4} = 0 $$

Approach $(0, 0)$ along $(t, \sqrt{t})$ as $t \to 0$ to get: $$ \lim_{t \to 0} \frac{t^2}{t^2 + t^2} = \frac{1}{2} $$

Thus, the limit doesn't exist.

In general, this question is a good resource on how to handle such problems. The technique I've used here is to find two paths that approach $(0, 0)$ but each gives a different limit.

edited Apr 13 '17 at 12:20

Community

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answered Mar 30 '13 at 00:07

Ayman Hourieh

39,603

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+1 Nice reference. For some illustrations, please see http://mathematica.stackexchange.com/questions/21544/finding-limits-in-several-variables. – whuber Mar 30 '13 at 00:13
@whuber Thanks for sharing. Pretty plots over there. :) – Ayman Hourieh Mar 30 '13 at 00:14
@AymanHourieh many many thanks!! I was just doing it with polar coordinates. And also, my profesor just do some simple exercises... :/ but many thanks! – Edwardo Mar 30 '13 at 00:30
@Edwardo Happy to help! – Ayman Hourieh Mar 30 '13 at 00:31
@AymanHourieh I was wondering, what happens if the numerator is x^2*y^2 ?? (This was the original exercise... It was a typo sorry) If I use the same method I have that L = 0 in both... now what? – Edwardo Mar 30 '13 at 01:13
@Edwardo In that case, the limit exists. It's too late to change the question and answer here. Please start a new question and I or someone else will help. – Ayman Hourieh Mar 30 '13 at 01:22
Idea of the solution: $$\frac{x^2y^2}{x^2+y^4} \le \frac{x^2y^2}{x^2} = y^2 \to 0$$. – Ayman Hourieh Mar 30 '13 at 01:23
@AymanHourieh Do you want to use the sandwich theorem? (it is squeeze the rigth name?) How I can find a function that is the lower bound effectively here?? – Edwardo Mar 30 '13 at 01:29
@Edwardo Squeeze theorem, yes. The lower bound is clearly $0$. – Ayman Hourieh Mar 30 '13 at 01:30
@AymanHourieh I think that I find it! – Edwardo Mar 30 '13 at 01:33
@AymanHourieh oh ok, I used then function (x^2*y^2)/(x^4 + y^4) and then I used polar coordinates on that and it tends to 0. Nevertheless, 0 is obvious. THANKS!!!! – Edwardo Mar 30 '13 at 01:35
@Edwardo Great! You're welcome. – Ayman Hourieh Mar 30 '13 at 01:37

Find $\lim\limits_{(x,y) \to(0,0)} \frac{xy^2}{ x^2 + y^4} $

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