Find a path to prove that the limit of $\frac{x^2}{x^2 + y^2 -x}$ does not exist I can't seem to find a path to show that:
$$\lim_{(x,y)\to(0,0)} \frac{x^2}{x^2 + y^2 -x}$$
does not exist.
I've already tried with $\alpha(t) = (t,0)$, $\beta(t) = (0,t)$, $\gamma(t) = (t,mt)$ and with some parabolas... they all led me to the limit being $0$ but this exercise says that there's no limit when approaching $(0,0)$. Hints? Thank you.
 A: So you want to find a path that gives a non-zero limit. Take a path where:
$$\lim_{(x,y)\to(0,0)} \frac{x^2}{x^2 + \color{blue}{y^2 -x}}$$the blue part is $0$, since then the limit clearly simplifies to $1$; so take $x=y^2$.
Edit: this comes down to s.harp's suggestion in the comment as well.
A: We can also try with parabolas with the axes exchanged, e.g. $x=y^2$:
$$\lim_{(x,y)\to (0,0),x=y^2}\frac{x^2}{x^2+y^2-x} = \lim_{y\to 0}\frac{y^4}{y^4+y^2-y^2}=1
$$
A: Other commenters have noted that the function value can be made to approach 1 along the parabolic path $x = y^2$.  In fact, the function's value along a path going to the origin can be made to approach any non-zero real number by choosing an elliptical or hyperbolic path instead.  If we look at the curve
$$
y^2 - x = \lambda x^2
$$
for an arbitrary constant $\lambda \neq 0$, then as we approach the origin along this curve, the value of $f(x,y)$ will approach
$$
\lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 + y^2 - x} = \lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 + \lambda x^2} = \frac{1}{1 + \lambda},
$$
and this latter fraction can be set equal to any $q \in \mathbb{R} \setminus \{0,1\}$ by choosing $\lambda = \frac{1}{q} - 1$.
The curves themselves can easily seen to be ellipses or hyperbolas by rearranging the above relation to read
$$
-\lambda \left( x + \frac{1}{2\lambda} \right)^2 + y^2 = -\frac{1}{4 \lambda},
$$
which is an ellipse when $\lambda < 0$ and is a hyperbola when $\lambda > 0$.
The parameterization of such a curve would be $(t, \sqrt{ \lambda t^2 + t})$;  note that in the case $\lambda < 0$, we would have to restrict the range of $t$ to $[0, -1/\lambda]$.
A: In fact this function, let's call it $f(x,y),$ is so badly behaved near $(0,0)$ that it maps every neighborhood of $(0,0)$ onto $\mathbb R.$
More precisey, set $E= \{(x,y): x\in [0,1], y = \sqrt {x-x^2}\}.$ Then the natural domain of $f$ is $U=\mathbb R^2\setminus E.$
Claim: For every $r>0,$ $f(U\cap D(0,r)) = \mathbb R.$
Here $D((0,0),r)$ is the open disc of radius $r$ centered at $(0,0).$
Proof of claim: This is actually easy. As you approach the point $(r/2, \sqrt {(r/2)-(r/2)^2}$ from below and and above you get the limits $-\infty,\infty$ respectively. Since $f$ is continuous in $U\cap D(0,r),$ and $U\cap D(0,r)$ is connected, $f(U\cap D(0,r))$ is a connected subset of $\mathbb R.$ There is no choice other than $f(U\cap D(0,r)) =\mathbb R,$ as claimed.
