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

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

  • $\begingroup$ In curves $y=ax^b$ are you assuming $x>0?$ $\endgroup$ – 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 $x\to 0^+$.

  • $\begingroup$ Good example, although it didn't answer the question because we are to assume limit $L$ on all those curves. $\endgroup$ – zhw. Oct 3 '17 at 19:03
  • $\begingroup$ 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 $\frac{f}{g}$. $\endgroup$ – Gio67 Oct 3 '17 at 19:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.