I need to prove that
$$\lim_{(x,y)\to(0,0)} \frac{xy(x^2-y^2)}{x+y} = 0. $$
I checked by approaching origin from all directions by substituting $$ y = mx $$ and doing $$\lim_{(x)\to(0)} \frac{xy(x^2-y^2)}{x+y} =\lim_{(x)\to(0)} \frac{mx^2(1-m^2)}{1+m^2} = 0 $$ for varying m.
So, the limit $L$ must be $0$.
I need to prove that given $$\sqrt{x^2+y^2} < d $$ , we can prove $$ \left|\frac{xy(x^2-y^2)}{x+y} - 0\right| < e $$ for some e, at all times.
I am not able to figure out how to simplify get rid of the denominator ( x + y ). Can someone help me solve this?
Since $x^2-y^2=(x-y)(x+y)$, you have$$\lim_{(x,y)\to(0,0)}\frac{xy(x^2-y^2)}{x+y}=\lim_{(x,y)\to(0,0)}xy(x-y)=0.$$