I have to prove that this limit doesn't exist.

$$\lim_{(x,y)\to(0,0)} \frac{xy}{x^2+y}$$

I tried this parametrization: $\begin{cases} x = t \\ y = mt^\alpha\end{cases}$

obtaining as result that the previous limit in this specific case would be equivalent to

$$\lim_{t\to0} \frac{mt}{t^{2-\alpha}+m}$$

which would be null for each value of $\alpha,m$.

Using a polar coordinate system doesn't seem effective too.

How do I prove that this doesn't exist?

  • 5
    $\begingroup$ Hint: note that along the path $y=-x^2$, the function does not exist at all. Along paths close to that, bad things should also happen. $\endgroup$
    – robjohn
    Mar 16, 2018 at 16:17
  • $\begingroup$ Damn you're right, how didn't I see that? $\endgroup$ Mar 16, 2018 at 16:20

4 Answers 4


Let $y=-x^2+x^4$. Then,


What happens now?

  • $\begingroup$ @Salahamam_Fatima Yes, I have edited before you sent the comment. $\endgroup$
    – Mark Viola
    Mar 16, 2018 at 16:20
  • $\begingroup$ that's a nice $\pm \infty$ $\endgroup$
    – user
    Mar 16, 2018 at 16:31

As an alternative to the solution by Mark Viola, note that

  • for $x=t \quad y=t \quad t\to 0 \implies \frac{xy}{x^2+y}=\frac{t^2}{t^2+t}=\frac{t}{t+1}\to0$

  • for $x=t \quad y=t^3-t^2 \quad t\to 0 \implies \frac{xy}{x^2+y}=\frac{t^4-t^3}{t^2+t^3-t^2}=\frac{t^4-t^3}{t^3}=t-1\to -1$

  • 3
    $\begingroup$ A question from a complete beginner: how did you select exactly this path among all the possible ones? Is there some way to spot it? $\endgroup$ Mar 16, 2018 at 16:32
  • 6
    $\begingroup$ @Bafforasta the first one is trivial (you can choose many of them, as also x=0 or y=0) for the second the trick is to try with a path for which the principal term of denominator cancel out (that's also the strategy used by Mark); in this case x^2+y=0 leads to $x=t$ and $y=-t^2$; then we add a higher order term to $y$ in such way that the denominator is not exactly =0, that is $y=-t^2+t^3$. This strategy often leads to different limits. $\endgroup$
    – user
    Mar 16, 2018 at 16:40

If $y=\frac {x^2}{x-1} $, the limit is

$$\lim_{(x,y)\to (0,0)}\frac {xy}{xy}=1$$

but if $x=\sqrt {y} $, it gives $$\lim_{(x,y)\to (0,0^+)}\frac {\sqrt {y}}{2}=0.$$ So, the limit cannot exist.


Let $$m=\frac{xy}{x^2+y}$$ Then $$m(x^2+y)=xy$$ so $$y=\frac{mx^2}{x-m}$$

For any $m$ we have an equation of a curve, which meets $(0,0)$ and on which the expression $\frac{xy}{x^2+y}$ is constant, equal $m$. This way there are infinitely many paths leading to the origin, each of which results in a different limit $m$.

Hence the limit does not exist.


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