I have solved the problem below, which gave me this question: How do i know when to use polar substitution?
If, instead of polar substitution, I directly set (x, y) = (0,0), I get division by zero. Therefore, I will instead use polar substitution. Is there anything else I should think about?
The problem.
How can the funtion be defined at the origin so that it becoms continuous at all points of the xy-plane?
$$f(x,y) = \frac{x^2+y^2-x^3y^3}{x^2+y^2}, \ (x,y)\neq(0,0)$$
My solution. $$\text{Polar substitution: } x=r\cos\theta,\ y=r\sin\theta,\ x^2+y^2=r^2\\ \begin{align} &\lim_{r\to0}\frac{r^2-r^3\cos^3\theta\, r^3\sin^3\theta}{r^2}\\ &\Rightarrow\lim_{r\to0} \frac{r^2(1-r\cos^3\theta\, r^3\sin^3\theta r)}{r^2(1)}\\ &\Rightarrow\lim_{r\rightarrow0} \frac{r^2(1-r\cos^3\theta\, r^3\sin^3\theta)}{r^2(1)}\\ &\Rightarrow\lim_{r\rightarrow0} \frac{1-r\cos^3\theta\, r^3\sin^3\theta}{1}\\ &\Rightarrow\lim_{r\rightarrow0} \frac{1-0\cos^3\theta\, 0^3\sin^3\theta}{1} \Rightarrow f(0,0)=1\\ \end{align} $$
If the polar substitution bounds the function between functions of $r$ that have the same limit, you can prove the intermediate function has the same limit by the squeeze theorem. In this case,$$f=1-r^4\cos^3\theta\sin^3\theta=1-\tfrac18r^4\sin^32\theta$$is bound between $1\mp\tfrac18r^4$. But even the looser, more obvious bounds $1\mp r^4$ prove the limit is $1$.