# On the existence of limits of multivariable rational functions

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

$$\lim_{(x,y) \to (0,0)} \frac{f(x,y)}{g(x,y)} = L?$$

• In curves $y=ax^b$ are you assuming $x>0?$ – zhw. Oct 2 '17 at 17:58

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 $x\to 0^+$.
• Good example, although it didn't answer the question because we are to assume limit $L$ on all those curves. – zhw. Oct 3 '17 at 19:03
• Thanks! The curve $y=x$ is not in the domain of the function $\frac{f}{g}$. When you calculate a limit $\lim_{x\to x_0}h(x)$ you only approach $x_0$ along points $x$ in the domain of $h$. So the limit is $L$ along all the curves which belong to the domain of $\frac{f}{g}$. – Gio67 Oct 3 '17 at 19:34