# How to calculate $\lim \limits_{(x,y)→(0,0)} \frac{(x^2+y^2)^2}{xy}$?

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?

• Wouldn't L'Hôpital be of help? Aug 27, 2020 at 9:19
• As you said : the limit won't exist, and your argument is correct.
– Surb
Aug 27, 2020 at 9:22
• 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.
– user
Aug 27, 2020 at 9:28
• 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. Sep 5, 2020 at 15:49

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 $$\lim_{r\to 0^+}\frac{2r^2}{\sin(2\theta)} = \lim_{r\to 0^+}\frac{2r^2}{\sin(2r^3)} = \lim_{r\to 0^+}\frac{1}{r}{\,\cdot\,}\frac{2r^3}{\sin(2r^3)} = \lim_{r\to 0^+}\frac{1}{r} = \infty$$