TITLE: On the existence of limits of multivariable rational functions
QUESTION [6 upvotes]: Standard limit-related counterexamples in multivariable calculus include limits like
$$\lim_{(x,y) \to (0,0)} \frac{2xy}{x^2 + y^2}$$
which tends to $0$ if the origin is approached along $x=0$ or $y=0$, but approaches $1$ if the origin is approached along the line $x=y$. This implies that the limit does not exist.
Indeed, there's rational functions for which the limit exists (and is the same) along all lines containing $(x_0, y_0)$, and yet the limit still fails to exist. For example, if we consider
$$\lim_{(x,y) \to (0,0)} \frac{2xy^2}{x^2 + y^4}$$
the limit is $0$ along lines of the form $y=\alpha x$ but $1$ along the curve $y^2 = x$.
I was wondering if we could have a more general counterexample of this sort.

Suppose $g(x,y)$ and $f(x,y)$ are two-variable polynomials defined on an open subset of $\mathbb{R}^2$ containing the origin such that $\lim_{(x,y) \to (0,0)} f(x,y) = \lim_{(x,y) \to (0,0)} g(x,y) = 0$.
In addition, suppose the rational function $$\frac{f(x,y)}{g(x,y)}$$
tends to some limit $L$ when $(0,0)$ is approached along curves of the form $y=\alpha x^{\beta}$ where $\alpha \in \mathbb{R}$ and $\beta>0$ (the limit $L$ is independent of the curve). Does it follow that
$$\lim_{(x,y) \to (0,0)} \frac{f(x,y)}{g(x,y)} = L?$$

REPLY [3 votes]: Take $f(x,y)=xy$ and $g(x,y)=x-y$. Then $$\frac{f(x,a x^b)}{g(x,a x^b)}=\frac{a x^{b+1}}{x-ax^b}.$$
Note that $a=1$ and $b=1$ are not allowed. If $b< 1$ you get 
$$\frac{f(x,a x^b)}{g(x,a x^b)}=\frac{a x^{b+1}}{x^b(x^{1-b}-a)}
=\frac{a x}{x^{1-b}-a}\to \frac{0}{0-a}=0.$$
If $b> 1$ you get 
$$\frac{f(x,a x^b)}{g(x,a x^b)}=\frac{a x^{b+1}}{x(1-ax^{b-1})}
=\frac{a x^b}{1-ax^{b-1}}\to \frac{0}{1-0}=0$$
and if $b=1$ you get 
$$\frac{f(x,a x)}{g(x,a x)}=\frac{a x^{2}}{x(1-a)}
=\frac{a x}{1-a}\to \frac{0}{1-a}=0$$
since $a\ne 1$.
However the limit does not exist since if you take $y=x+x^3$ you get
$$\frac{f(x,x+x^3)}{g(x,x+x^3)}=\frac{x^2+x^4}{x^3}
=\frac{1+x^2}{x}\to \infty$$
as $x\to 0^+$.