Find $\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4}$

Question

Find $\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4}$

I don't know how to approach it without using polar coordinates. Any hints?

Answer 1

Hint: Use inequality for arithmetic and geometric mean, i.e. $x^4 + y^4 \geqslant 2 x^2 y^2 \geqslant x^2 y^2$.

Answer 2

$(x^2-y^2)^2 = x^4+y^4 -2x^2y^2\ge 0.$

$\rightarrow$

$\dfrac{x^2y^2}{x^4+y^4}\le 1/2$.

Hence : $0\le |\dfrac{x^3y^2}{x^4+y^4}| \le (1/2)|x| \le$

$(1/2)\sqrt{x^2} \le (1/2)\sqrt{x^2+y^2}.$

$\epsilon$ given choose $\delta =2 \epsilon.$