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

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).

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.
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.