I have to prove that

$$\lim_{(x,y)->(0,0)} \frac{xy}{2x-y}$$ doesn't exist.

I have tried to use these restrictions:

  • $x=0; y=0; y=x; y=mx; y=mx+q; y=ax^2;y=ax^2+bx+c; y=1/x, y=1/x^2,..$

and for each of them, I have done the related limit. But I have always obtained 0. I know that I if want to prove that the limit doesn't exist, I have to find two restrictions that have different values for their limits.

Any suggestions? Many thanks

  • $\begingroup$ Try $y=2x~~~~$. $\endgroup$ – vadim123 Mar 10 '15 at 18:59
  • 1
    $\begingroup$ @vadim123 : This line is outside the domain of the function. $\endgroup$ – Tryss Mar 10 '15 at 19:00
  • $\begingroup$ @vadim123 That path is undefined for all $x$. $\endgroup$ – Alice Ryhl Mar 10 '15 at 19:00
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    $\begingroup$ $y=2x-x^2$ I guess $\endgroup$ – Brian Ding Mar 10 '15 at 19:02
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    $\begingroup$ @sunrise, what I tried is first $y=x$,$y=x^2$ as you. Notice that $(x,y)\rightarrow (0,0)$ and the denominator is $2x-y$, I think $y=2x+sth$ would be a good one and the easiest 'sth' would be $x^2$. $\endgroup$ – Brian Ding Mar 10 '15 at 20:30

$$ \lim_{(x,y)->(0,0)} \frac{xy}{2x-y} $$ Let $y=ax^2+bx+c$ then we get $$ \lim_{(x,y)->(0,0)} \frac{ax^3+bx^2+cx}{2x-ax^2-bx-c} $$ We can see that for it to have a nonzero numerator, we must cancel an $x$. To do that we must set $c=0$ giving. $$ \begin{align} &\lim_{(x,y)->(0,0)} \frac{ax^3+bx^2}{2x-ax^2-bx}\\ =\:&\lim_{(x,y)->(0,0)} \frac{ax^2+bx}{2-ax-b} \end{align} $$ We now obtain the same problem as before, we must cancel an $x$, to do that we must set $b=2$. $$ \begin{align} &\lim_{(x,y)->(0,0)} \frac{ax^3+2x^2}{2-ax^2-2}\\ =\:&\lim_{(x,y)->(0,0)} \frac{ax^3+2x^2}{-ax^2}\\ =\:&\lim_{(x,y)->(0,0)} \frac{ax+2}{-a} \end{align} $$ Take the limit and obtain. $$ \lim_{(x,y)->(0,0)} \frac{ax+2}{-a}=\frac{-2}a $$ The limit is different for every $a\in\mathbb R$, so the original limit is undefined.

  • 1
    $\begingroup$ or just the limit is different for every $a$ :-) $\endgroup$ – Ant Mar 10 '15 at 19:44
  • $\begingroup$ @Ant yeah, this is more nice $\endgroup$ – Alice Ryhl Mar 10 '15 at 19:46
  • $\begingroup$ oh, it's very interesting! I didn't think to set some values for b and c... so many thanks! $\endgroup$ – sunrise Mar 10 '15 at 19:56

$$\lim\limits_{(x,y)\to (0,0)} \frac{xy}{2x-y}$$ Using polar coordinates, we have $$\lim\limits_{r\to 0^+} \frac{r^2\cos\phi \sin\phi}{2r\cos\phi-r\sin\phi}$$ $$=\lim\limits_{r\to 0^+} \frac{r\cos\phi \sin\phi}{2\cos\phi-\sin\phi}$$ Now let's attempt to find a bound that is independent of $\phi$, $$\left|\frac{\cos\phi \sin\phi}{2\cos\phi-\sin\phi}\right|\leq \left|\frac{\frac12}{2\cos\phi-\sin\phi}\right| $$ $$\left|\frac{\cos\phi \sin\phi}{2\cos\phi-\sin\phi}\right|\leq \left|\frac{1}{4\cos\phi-2\sin\phi}\right| $$ The right hand side is unbounded which implies that the limit is dependent on $\phi$. Therefore $$\lim\limits_{(x,y)\to (0,0)} \frac{xy}{2x-y}\Rightarrow \mbox{does not exist}$$


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