3
$\begingroup$

Im trying to find: $\lim\limits_{(x,y)\to(0,0)}\frac{x^2}{x^2+y^2-x}$

  • If I take the path $x=y$ that limit is $0$:

$\lim\limits_{(x,y)\to(0,0)}\frac{y^2}{y^2+y^2-y}=\lim\limits_{(x,y)\to(0,0)}\frac{y}{2y-1}=0$

  • If I take the path $x=y^2$ that limit is $1$:

$\lim\limits_{(x,y)\to(0,0)}\frac{y^4}{y^4+y^2-y^2}=\lim\limits_{(x,y)\to(0,0)}\frac{y^4}{y^4}=1$

So the limit doesn't exist.

But wolframalpha says that limit is $0$.

I tried polar coordinates and I get: $\lim\limits_{r\to0}\frac{rcos^2\alpha}{r-cos\alpha}$

  • If $cos\alpha \neq 0$ that limit is $0$.

  • If $cos\alpha = 0$ that means that $x = 0$. Then:

$\lim\limits_{(x,y)\to(0,0)}\frac{x^2}{x^2+y^2-x}$ = $\lim\limits_{(x,y)\to(0,0)}\frac{0}{y^2}=0$

So the limit is $0$.

Wich one is true? What is correct and what is wrong? (Excuse me for my english).

$\endgroup$
0
$\begingroup$

When you say, "If $\cos\alpha\neq0$", you are treating $\cos \alpha$ as a constant, which is equivalent to treating $y/x$ as a constant. So this means that if you approach $(x,y)=(0,0)$ along a path where $y/x$ is a constant, the limit is 0.

That is consistent with your finding in the first approach.

$\endgroup$
-2
$\begingroup$

It sounds like asking to compute the limit of $x/y$ as both $x$ and $y$ tend to 0. The limit depends on which variable tends to 0 faster. The limit under a given path exists, but not the general limit, in the same way as a right-side limit and left-side limit for a univariate function may both exist, but differ.

$\endgroup$
  • 1
    $\begingroup$ No, that's not what OP is asking. $\endgroup$ – Stefan Mesken Nov 4 '16 at 2:41

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.