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 $1$

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 $0$
$\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2 -x^3 y^3}{x^2 +y^2} = 1-a = 1$
