For what positive integers a, b, c, and d, where $c \le d$, does the limit as $\begin{pmatrix} x\\ y\\ \end{pmatrix} \to \begin{pmatrix} 0\\ 0\\ \end{pmatrix}$ of

$$\frac{x^ay^b}{x^{2c}+y^{2d}} \quad \text{exist?}$$

This is a minor modification of problem 1.5.24 on p 104 of Hubbard & Hubbard's Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach (5th edition, ISBN 9780971576681, pub by Matrix Editions). H & H said a, b, c, and d are non-negative integers and didn't specify $\,$ $c\le d$ , but $c\le d$ may be assumed without loss of generality, and if we allow any exponent to be 0 there are lots of trivial cases to deal with, all of which I can solve.

This is what I've proven so far:

1) $\,$If $\,a \gt c \,$ and $\, b \gt d\,$ the limit is 0.$\,$ But I don't know what happens in general if $ \,a \le c \,$ or $ \,b \le d$.

2) $\,$ If $a \gt 2c \,$ or $\, b \gt 2d\,$ the limit is 0. $\,$ But I don't what happens if $\, a \le 2c \,$ and $\,b \le 2d$.

3) $\,$ If $ \,a+b \le 2c \,$ the limit does not exist, and if $\, a+b \gt 2d \,$ the limit is 0. $\,$ But I don't know what $\quad$ happens if $\, 2c \lt a+b \le 2d$.

4) If $\,ad + cb \le 2cd \,$ the limit does not exist, which implies that if $\,a \le c \,$ and $\, b \le d \,$ then the limit $\quad$ also does not exist. $\,$But I don't know what happens if $ \,ad + cb > 2cd$.

Since at this point in the book H & H haven't even defined partial derivatives, I'm hoping for a proof that requires nothing fancier than AP BC calculus and the basic limit definitions and properties of limits for functions from $R^n$ into $R^m$. But any proof or help will be greatly appreciated.



1 Answer 1


HINT: Let $x^c = r\cos\theta$ and $y^d = r\sin\theta$.

(Comment: This is one of my favorite questions. You don't need $a,b,c,d$ to be integers, either. Just positive real numbers.)


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