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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).

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2 Answers 2

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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.

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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.

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    $\begingroup$ No, that's not what OP is asking. $\endgroup$ Nov 4, 2016 at 2:41

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