The wolframalpha gives the answer $0$:

Wolframalpha culculation

I tried like this:

Let $x=r\cos(\theta)$ and $y=r\sin(\theta)$,then:

$\lim \limits_{(x,y)→(0,0)} \frac{(x^2+y^2)^2}{xy}$ = $\lim \limits_{r→0} \frac{r^4}{r^2 \sin(\theta)\cos(\theta)}$ = $\lim \limits_{r→0} \frac{2r^2}{\sin(2\theta)} =0$

but it seems wrong when $\theta =0$ and the limit $\lim \limits_{r→0} \frac{2r^2}{\sin(2\theta)}$ may not exist.

So how to calculate the limit?

  • $\begingroup$ Wouldn't L'Hôpital be of help? $\endgroup$
    – Déjà vu
    Aug 27, 2020 at 9:19
  • 2
    $\begingroup$ As you said : the limit won't exist, and your argument is correct. $\endgroup$
    – Surb
    Aug 27, 2020 at 9:22
  • 1
    $\begingroup$ You can't take $\theta=0$, we need to show at least two different "paths" with different limits exist. Your one is a good qualitative argument but it is not a proof. $\endgroup$
    – user
    Aug 27, 2020 at 9:28
  • $\begingroup$ You can choose a simple path for this $y = mx$ so that it has a dependency on x and as $x \to 0$ we have zero as the answer. $\endgroup$ Sep 5, 2020 at 15:49

2 Answers 2


We have that for $x=y=t\to 0$

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

but for $x=t\to 0$ and $y=t^3$

$$\frac{(x^2+y^2)^2}{xy}=\frac{(t^2+t^6)^2}{t^{4}}=1+2t^4+t^8 \to 1 $$

therefore the limit doesn't exist.


Note that the $x$ and $y$ axes are not in the domain of the given function, hence for $r$ sufficiently small, $\sin(2\theta)\ne 0$.

Nevertheless, to show that the limit does not exist, if we let $(x,y)$ approach $(0,0)$ along the curve $\theta=r^3$, we get $$ \lim_{r\to 0^+}\frac{2r^2}{\sin(2\theta)} = \lim_{r\to 0^+}\frac{2r^2}{\sin(2r^3)} = \lim_{r\to 0^+}\frac{1}{r}{\,\cdot\,}\frac{2r^3}{\sin(2r^3)} = \lim_{r\to 0^+}\frac{1}{r} = \infty $$


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