# An interesting problem on a function in two variables

One of the starkest differences between dealing with differentiation of a multi-variable function and a single variable function is that there are now an infinite number of ways to take a limit, and it becomes much more difficult to show that the derivative exists at a point by checking that derivatives exist along all possible paths. With functions in two variables, for example, one can usually play around with $x$ and $y$ separately to try to get different values at a limit point. Recently I ran across a problem that I found striking.

Let $a,b,c,d$ be positive real numbers and let

$\displaystyle f(x,y) = \frac{|x|^a |y|^b}{|x|^c + |y|^d}$.

Prove that $\lim_{(x,y) \rightarrow (0,0)} f(x,y) = 0$ if $\displaystyle \frac{a}{c} + \frac{b}{d} > 1$ and that the limit does not exist if $\frac{a}{c} + \frac{b}{d} \leq 1.$

I found this particularly interesting because it is not at all clear why the quantity $\displaystyle \frac{a}{c} + \frac{b}{d}$ should control the existence of the limit at $(0,0)$. The proof, I thought, was quite clever.

In retrospect the case $a/c + b/d \leq 1$ is easier, as we just have to exhibit a path for which the limit as $(x,y)$ approaches $(0,0)$ along the path is non-zero. Indeed, trivially we have $\lim_{(x,0) \rightarrow (0,0)} f(x,0) = 0$ for any positive values of $a,b,c,d$, and so this suffices. Now we choose the path where $x,y > 0, y = x^{c/d}$. Then we have

$\displaystyle f(x,y) = \frac{x^a x^{bc/d}}{x^c + x^c} = \frac{1}{2}x^{c\left(\frac{a}{c} + \frac{b}{d} - 1\right)}$

and this limit is plainly $1/2$ if $a/c + b/d = 1$ and $+ \infty$ if $a/c + b/d < 1$.

The other situation is much trickier. We will require two ad-hoc inequalities.

$\displaystyle |x| = (|x|^c|)^{1/c} \leq (|x|^c + |y|^d)^{1/c}$,

$\displaystyle |y| = (|y|^d)^{1/d} \leq (|x|^c + |y|^d)^{1/d}.$

Now we have

$\displaystyle |x|^a |y|^b \leq (|x|^c + |y|^d)^{a/c} (|x|^c + |y|^d)^{b/d},$

whence

$0 \leq f(x,y) \leq (|x|^c + |y|^d)^{a/c + b/d - 1},$

and if $a/c + b/d - 1 > 0$ then the right hand side tends to $0$ as $(x,y) \rightarrow (0,0)$, thus establishing our proposition.

along this path.