Limits and Continuity Proof

Question

Let $f(x,y) = \frac{x^ay^b}{x^2+y^2}$

Show that $\lim(x,y)->(0,0)$ $f(x,y) = 0$ if a+b>2 and does not exist if a+b<2.

Should I set a value for a and b (and use a case by case basis) to prove this or is there another way?

Answer 1

Transforming to polar coordinates with $x=\rho \cos \phi$ and $y=\rho \sin \phi$ yields

$$\frac{x^ay^b}{x^2+y^2}=\rho^{a+b-2}\cos ^a\phi \sin^b \phi \tag 1$$

As $(x,y)\to (0,0)$, $\rho \to 0$. Therefore, the right-hand side of $(1)$ goes to $0$ whenever $a+b-2>0$ since the sine and cosine functions are bounded (in absolute value) by $1$.

If $a+b-2<0$, then the limit is $0$ when $\phi =0$ and the limit is $+ \infty$ if, say, $\phi =\pi/4$. Inasmuch as the limit much be independent on its path, then the limit is undefined for $a+b-2<0$.