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I have to prove that the limit of the function $\frac{x^2}{x^2+y^2}$ as $x$ approaches infinity and as y approaches infinity does not exist.

I thought about finding the side limits, and if they are not equal, bam! I have solved it. But what should I take as side limits here? $+$ and $-$ infinity? Thank you in advance :)

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    $\begingroup$ For two-dimensional limits, there are not just 2 sides. $\endgroup$ – GEdgar Mar 18 '13 at 21:05
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    $\begingroup$ Try looking as $x$ goes to infinity, then as $y$ goes to infinity, and vice versa. $\endgroup$ – ferson2020 Mar 18 '13 at 21:05
  • $\begingroup$ @rschwieb, I don't think that's possible: both variables must go infinity, this or that way or rate, but both at the same time. $\endgroup$ – DonAntonio Mar 18 '13 at 21:17
  • $\begingroup$ But the paths you gave @rschwieb are forbidden in this case as both variables must go to infinity. If you had chosen $\,y=x^2\;,\;x=3y-8\,$ or something else then fine, but not when you fix one variable that should be running to infinity $\endgroup$ – DonAntonio Mar 18 '13 at 21:27
  • $\begingroup$ @DonAntonio Yes I see what you mean. I had some picture in my head about observing different paths "away from the origin", but I think you're right. $\endgroup$ – rschwieb Mar 18 '13 at 21:28
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$$(1)\;\;\;\;y=x\;\;\;\;\Longrightarrow\;\;\; \lim_{x,y\to\infty}\frac{x^2}{x^2+y^2}=\lim_{x\to\infty}\frac{x^2}{2x^2}=\frac{1}{2}$$

$$(2)\;\;\;\;\;\;\;\;\;\;\;\;\;y=x^2\;\;\;\;\Longrightarrow\;\;\; \lim_{x\to\infty}\frac{x^2}{x^2+x^4}=0$$

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    $\begingroup$ Gracias amigo :) $\endgroup$ – asdas Mar 18 '13 at 22:15
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ok at first we say we observe $x\to \infty$ and $x=y$ than the limit is the same as $$\lim_{x\to \infty}\frac{x^2}{2x^2}=\frac{1}{2}$$ Now we look at $x=2y$ than the limit is $$\lim_{y \to \infty} \frac{4y^2}{4y^2 +y^2}=\frac{4}{5}$$ but the limits must be same so they don't exists.

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Let $x=r\cos\theta$ and $y=r\sin\theta$. Then $$\frac{x^2}{x^2+y^2}=\cos^2\theta$$ if $(x,y)\ne (0,0)$. Thus even for very large positive $x$ and $y$, the function $\frac{x^2}{x^2+y^2}$ can take on any value in the interval $(0,1)$.

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The point you need to know, and glean from excellent examples posted, is to check the limit of your function as $(x, y) \to \infty$ along two or more different curves.

You are working in two dimensions, so the choice of curves you can test can is not one dimensional lines. Put, say, $y = x$, and $y = x^2$, perhaps even $y = x^3$.

Then, write the function as a function of $x$ (replace $y$ in the function $f(x)$ for each curve, depending on the curve $y$) and evaluate limit of the function as $x \to \infty$.

E.g., with your function, if we let $y = x$, then substitute $x$ whenever $y$ appears in you function, and evaluate: $$\lim_{x\to \infty} \frac{x^2}{2x^2} = 1/2\tag{1}$$

And if $y = x^{-2}$, we substitute $x^{-2}$ for $y$ and then evaluate $$\lim_{x\to \infty} \frac{x^2}{1 + x^2} = 1\tag{2}$$

If the limits as $x\to \infty$ of the functions defined in terms of different curves, are not equal, as we have with $(1), (2)$, you know the limit does not exist.

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I like Andre Nicolas' answer and mine is similar. Since the degree of the terms in the denominator are of equal degree, let $$y=ax.$$ Then we have $$1/(1+a^2).$$

Suppose you have $$x^2 +y^4$$ in the denominator, with $$y^4$$ in the numerator. Then let $$x=ay^2.$$ So you see that this approach is generalizable.

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