Proving that $\lim_{(x,y) \to (0,0)} (x^2 +y^2 -x^3 y^3)/(x^2 +y^2) =1$ How can I go about proving that $$\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2 -x^3 y^3}{x^2 +y^2} = 1 ?$$
I checked some lines along $x, y$ and $x=y$ and it all gave $1$
 A: An alternative solution without polar coordinate:
$$\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2 -x^3 y^3}{x^2 +y^2} = \lim_{(x,y) \to (0,0)} 1- \frac{x^3 y^3}{x^2 +y^2}$$
To show $\lim_{(x,y) \to (0,0)} \frac{x^3 y^3}{x^2 +y^2} = 0$ is equivalent to show $\lim_{(x,y) \to (0,0)} |\frac{x^3 y^3}{x^2 +y^2}| = 0 $:
Since $$0 \leq |\frac{x^3 y^3}{x^2 +y^2}| = \frac{|x|^3 |y|^3}{x^2 +y^2}   \leq^{[1]} \frac{|x|^3 |y|^3}{2|x||y|} = \frac{|x|^2 |y|^2}{2} $$
Where the inequality $[1]$ is with AM-GM inequality.
By Squeeze Theorem, $ 0 \leq \lim_{(x,y) \to (0,0)} |\frac{x^3 y^3}{x^2 +y^2}| \leq \lim_{(x,y) \to (0,0)}\frac{|x|^2 |y|^2}{2} = 0$.
So $\lim_{(x,y) \to (0,0)} |\frac{x^3 y^3}{x^2 +y^2}| = 0 $ which means $\lim_{(x,y) \to (0,0)} \frac{x^3 y^3}{x^2 +y^2} = 0 $
A: In polar coordinates, this can be written as
$$1 - r^4 \cos^3 \theta \sin^3 \theta$$
and $(x, y) \to (0, 0)$ is equivalent to taking $r \to 0$.
A: The first step,
$\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2 -x^3 y^3}{x^2 +y^2} = \lim_{(x,y) \to (0,0)} 1- \frac{x^3 y^3}{x^2 +y^2}$
Now study this and find $a$
$1-a=1-\lim_{(x,y) \to (0,0)} \frac{x^3 y^3}{x^2 +y^2}$
$a=\lim_{(x,y) \to (0,0)}  \frac{x^3 y^3}{x^2 +y^2}$
$\frac {1}{a}=\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2}{x^3y^3} $
$=\lim_{(x,y) \to (0,0)}  \frac{x^2}{x^3 y^3}+\frac{y^2}{x^3 y^3}=\lim_{(x,y) \to (0,0)}  \frac{1}{xy^3}+\frac {1}{x^3y}=\infty$
Then $a$ must be $0$
$\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2 -x^3 y^3}{x^2 +y^2} = 1-a = 1$
