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Hi i'm having this doubt about this multivariable limit

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

If I use these two paths like let $x = y^3$ and $y=x$ I get that the limit is going to be 0, however if I add this other path where $x=\sqrt y$ the limit is going to be 1, and that tells me the limit doesn't exist, but what if I just used the first ones? while doing the proof by definition shouldn't this be refuted? cause I don't know if i'm applying this the wrong way and ending up with a valid limit when I shouldn't, how do you guys know exactly which paths you can use to say that the limit doesn't exists? is there like a rule or something similar? Thanks in advance!

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in multivariable functions you can say that a limit exist proving it by definition. You only use different paths to see if a limit doesn't exist (case that you get different values), or to find a candidate to be the limit and use it to proving that by definition.

You don't have a way to see exactly which paths you have to use. You can use the lines x=0, y=mx and then all depends on your creativity.

Sometimes can be useful to plot some level sets of the function, and analize if exist a trivial path which aproximates the point in a special way that can give a diferent value of the limit.

I hope it had been helpful.

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  • $\begingroup$ Exactly, in this case i used the first two paths as i earlier mentioned I got 0, then by definition that's the value i used then I got to the point where the absolute value of the the function is < than Epsilon and I thought I proved it, but then I noticed afterwards because a friend told me if you used that path where $x = \sqrt y$ the limit doesn't exists cause is different as the other two paths I already used, but then again the main question is does the definition tells me if it really exists? even wolfram alpha says this limit is 0 however in leithold's 7th edition says it doesn't exists $\endgroup$ – NeptaliD Sep 2 '16 at 3:01
  • $\begingroup$ Yes, limit by definition always work. Maybe you do something wrong when you try to proof it? I try to solve it and using the path x=sqrt(y) the limit goes to 1, but using the path y=mx goes to 0. So, the limit doesn't exist. Don't trust always on Wolfram Alpha. $\endgroup$ – DN_Euler Sep 2 '16 at 3:12
  • $\begingroup$ Alright thanks so much yeah i'm gonna try and see what's the failure with the definition I tried $\endgroup$ – NeptaliD Sep 2 '16 at 3:15
  • $\begingroup$ Excellent, good luck! $\endgroup$ – DN_Euler Sep 2 '16 at 3:17
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The statement that the limit exists means that for all neighborhoods $N_\varepsilon L$ there is a neighborhood $M_\delta (0,0)$ such that whenever $x \in M_\delta(0,0)$, it follows that $f(x) \in N_\varepsilon L$.

Thus, if you can find two paths that give different limits, the limit cannot exist since our condition about being in some $\delta$-neighborhood implying proximity to $L$ is not satisfied.

Consider the path $(x, x^2)$: $$ \lim_{(x,x^2)\to(0,0)} \frac{2x^2x^2}{x^4+(x^2)^2} = \lim_{(x,x^2)\to(0,0)} \frac{2x^4}{2x^4} = 1. $$

Consider the path $(x,0)$, that is, approaching along the $x$-axis: $$ \lim_{(x,0)\to(0,0)} \frac{2x^2\cdot 0}{x^4+0} = \lim_{(x,0)\to(0,0)} 0 = 0. $$

Hence, the limit does not exist. In general, if we evaluate the limit of a function $f$ along a certain path $p_1$ and determine the limit along this path is, say $L'$, then that is all we have determined. However, to say that the limit of the function is some particular value $L$ means that it is the value we achieve if we approach along every path. Thus, we have not shown the limit is anything until we have shown that the limiting value is the same for all paths.

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  • $\begingroup$ I get that, however how do you prove that the limit doesn't exist by definition? i don't see what i did wrong while proving it was 0, and it is a wrong answer $\endgroup$ – NeptaliD Sep 2 '16 at 4:28
  • $\begingroup$ @NeptaliD, I have updated my answer to address that question. $\endgroup$ – Alex Ortiz Sep 2 '16 at 4:34
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The following is the general result on this:

Lemma For positive $\alpha, \beta, \gamma, \delta, \epsilon$ The limit $$\lim_{(x,y)\to (0,o)} \frac{|x|^\alpha |y|^\beta}{(|x|^\gamma+|y|^\delta)^\epsilon} = \left\{ \begin{array}{lr} 0 & \mbox{ if } \frac{\alpha}{\gamma}+\frac{\beta}{\delta} >\epsilon \\ \mbox{DNE} & \mbox{ if } \frac{\alpha}{\gamma}+\frac{\beta}{\delta} \leq \epsilon \end{array}\right.$$

Proof:

$$|x|^\alpha |y|^\beta= (|x|^\gamma)^{\frac{\alpha}{\gamma}}(|y|^\delta)^{\frac{\beta}{\delta}}\leq (|x|^\gamma+|y|^\delta)^{\frac{\alpha}{\gamma}}(|x|^\gamma+|y|^\delta)^{\frac{\beta}{\delta}} \\ \frac{|x|^\alpha |y|^\beta}{(|x|^\gamma+|y|^\delta)^\epsilon} \leq (|x|^\gamma+|y|^\delta)^{ \frac{\alpha}{\gamma}+\frac{\beta}{\delta} -\epsilon}$$

which proves the first part.

For the second, pick the curve paramatrised by $|x|^\gamma=|y|^\delta=t$ and calculate the limit along this curve.


In your problem, you need to decide if $\frac{2}{4}+\frac{1}{2}$ is $>$ or $\leq $ than 1.

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