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.

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