6
$\begingroup$

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?

$\endgroup$
  • 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 '18 at 16:17
  • $\begingroup$ Damn you're right, how didn't I see that? $\endgroup$ – Baffo rasta Mar 16 '18 at 16:20
12
$\begingroup$

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

$$\frac{xy}{x^2+y}=-x^{-1}+x$$

What happens now?

$\endgroup$
  • $\begingroup$ @Salahamam_Fatima Yes, I have edited before you sent the comment. $\endgroup$ – Mark Viola Mar 16 '18 at 16:20
  • $\begingroup$ that's a nice $\pm \infty$ $\endgroup$ – user Mar 16 '18 at 16:31
7
$\begingroup$

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

$\endgroup$
  • 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? $\endgroup$ – Baffo rasta Mar 16 '18 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 '18 at 16:40
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.