Proving that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2y^3}{x^2+y^2}=0$

Question

I need to show that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2y^3}{x^2+y^2}=0$. My attempted method was showing that as the distance between $(x,y)$ and $(0,0)$ approaches $0$, so does the function in the limit. I tried to show $\lvert{\frac{x^2y^3}{x^2+y^2}}\rvert \leq \sqrt{x^2+y^2}$ for $(x,y) \neq 0$. I expanded the terms to get $\frac{x^2\lvert y^3\rvert}{x^2+y^2} \leq \frac{(x^2+y^2)\sqrt{x^2+y^2}}{x^2+y^2}$, or $(x^2+y^2)^\frac{3}{2}\geq x^2\lvert y^3\rvert$. I am unsure of how to prove this and whether this method is valid or would work in showing the existence of the above limit at all.

Answer 1

Hint: Where defined, $\frac{1}{x^2 + y^2}\leq \frac{1}{y^2}$

Answer 2

A common approach, when you see squares as in the denominator, is conversion to polar coordinates: $x=r\cos t$, $y=r\sin t$, then

$$ \lim_{(x,y)\to(0,0)}\frac{x^2y^3}{x^2+y^2} = \lim_{r\to 0} \frac{(r\cos t)^2(r\sin t)^3}{(r\cos t)^2 + (r\sin t)^2} = \lim_{r \to 0} \frac{r^5 \cos^2 t \sin^3 t}{r^2(1)} = \cos^2 t\sin^3 t \lim_{r\to 0}r^3 = 0 $$

Answer 3

Note : $x^2+y^2 \ge |2xy|$.

Let $ |x^2+y^2|^{1/2} \lt \delta, $

Choose $\epsilon = \delta^3 /4$.

$(x,y) \ne (0,0)$.

$|\dfrac{x^2y^3}{x^2+y^2}| \le$

$ (1/4)|\dfrac{y(x^2+y^2)^2}{x^2+y^2}| =$

$ (1/4)|y||(x^2+y^2)| \lt $

$ (1/4) (\delta)(\delta)^2 = \epsilon$.

Used: $ |y| \le (x^2+y^2)^{1/2} \lt \delta $.