I'm trying to compute the following limit:

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

I think that the limit is actually path dependent, thus does not exist.

If we are looking on the path $(x,y)=(t^2,k^2 t)$ for some $k\in \Bbb R$ we get that $$\lim_{(x,y)\to(\infty,\infty)}\frac{x^2+y^2}{x^2+y^4}=\lim_{t\to\infty}\frac{t^4+k^2t^2}{t^4+k^8t^4}=\lim_{t\to\infty}\frac{1+\frac{k^2}{t^2}}{1+k^8}=\frac{1}{1+k^8}$$ Hence, the limit is path dependent, so it does not exist.

W|A claims that the limit is 0. What is wrong with my reasoning?

  • 2
    $\begingroup$ Someone posted an incorrect solution earlier using polar coordinates. He deleted it, but just in case he still wants to know why it's wrong: @Christian The problem with your method is that $\lim_{(x,y)\to(\infty,\infty)}\frac{x^2+y^2}{x^2+y^4} \neq \lim_{r\to \infty} \frac{1}{\cos^2(\phi) + r^2 \sin^4(\phi) } = 0, \phi \in (0, \frac{\pi}{4}]$. This is because your $x$ and $y$ can have differing values for $\phi \in (0, \frac{\pi}{4}]$ as they go to infinity, but your calculation of your limit assumes a constant positive value of $\phi \in (0, \frac{\pi}{4}]$. $\endgroup$
    – Ilham
    May 10 '15 at 9:47

Your way looks fine.

Wolfram alpha is not very reliable with multivariable limits, as far as I recall

  • $\begingroup$ Thank you, I think that I will contact W|A so they will fix it. $\endgroup$
    – Galc127
    May 10 '15 at 10:25

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