Question

Multivariable limit question.

Answer 1

\[\lim_{(x,y) \rightarrow (0,0)} \frac{ 2xy^2 }{ x^2+y^4 }\]

How do I approach this to turn it into limits I can do with L'hopital's rule or something.

Answer 2

I suppose I need to plug in y=mx somehow actually I might be able to figure it out...

Answer 3

Do you know about partial derivatives?

Answer 4

Yeah, totally.

Answer 5

Pretend that \(y\) is a constant and try to evaluate it.

Answer 6

Actually, scratch that you can't do what I said because you wouldn't get an indeterminate form necessarily....

Answer 7

No, the limit should be 0 at that point. I guess I get confused because it would seem to me that I know how to show a limit doesn't exist, but in this case I want to show that if there was a point at 0,0 on this graph then it would be continuous here.

Answer 8

Jumbled words is what that looks like I just typed hopefully that makes sense lol.

Answer 9

Is this: \[
\lim_{(x,y) \rightarrow (0,0)} \frac{ 2xy^2 }{ x^2+y^4 }
\]Different than this: \[
\lim_{x\to 0}\; \lim_{y\to 0} \frac{ 2xy^2 }{ x^2+y^4 }
\]Can't we do try to evaluate the inner limit first and then the outer one?

Answer 10

I don't know, possibly. Since if you evaluate either x or y first then the limit becomes 0 which is the answer, but it feels wrong.

Answer 11

Umm, if you approach with \(x\) being negative, then you get \(-\infty\) and if you approach with x being positive you get \(\infty\) so does this limit actually exist?

Answer 12

try y=mx, you get .. m^2x^3/(x^2 + m^4x^4) = xm^2/(1 + m^4x^2) ... which goes to zero as x->0

Answer 13

shows that your limit goes to zero on every curve y=mx

Answer 14

Alright, yeah what you are both saying makes sense to me.

But experimentX, does this mean that the limit exists or should I keep trying different curves to try to find one that doesn't work? How do I know, basically.

Answer 15

I can't say exactly, but if the bottom x^2 were x^3, it was sufficient to prove that limit did not exist. I am still searching for better method.

Answer 16

there isn't much worked out example on my book ... just bunch of silly questions.

Answer 17

Yeah same here with my book, thanks for the help.

Answer 18

Even single limits are not easy to predict...

Answer 19

maybe we could consider following cases. y=f(x), be the a curve
case 1, lim x->0 f(x)/x = 1
case 2, lim x-> 0 f(x)/x = infty
case 2, lim x-> 0 f(x)/x = 0
the single limit is zero, as well as iterated limit ... but it's not enough to show the double limit is zero :(((

Answer 20

I think the double limit can be done well if you do: \[
\lim_{(x,y)\to (a,b)} \to \lim_{x\to a}\;\lim_{y\to f(x)}
\]Where \(f(a)=b\)

Answer 21

hmm ,,, yeah, but no need to write y->f(x),
x-> a will suffice.

Answer 22

\[ \lim_{(x,y)\to (a,b)} g(x,y) \to \lim_{x\to a} g(x, f(x))\]

Answer 23

the above theorem is essentially what wio originally suggested, and gives value 0. By defining curve is x,y plane,
All cases zero which generalizes the all possible curve.