3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 5
    $\begingroup$ $y=2x-x^2$ I guess $\endgroup$ – Brian Ding Mar 10 '15 at 19:02
  • 1
    $\begingroup$ @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
4
$\begingroup$

$$ \lim_{(x,y)->(0,0)} \frac{xy}{2x-y} $$ 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.

$\endgroup$
  • 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
1
$\begingroup$

$$\lim\limits_{(x,y)\to (0,0)} \frac{xy}{2x-y}$$ 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 $$\lim\limits_{(x,y)\to (0,0)} \frac{xy}{2x-y}\Rightarrow \mbox{does not exist}$$

$\endgroup$

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.