Prove, using first principles, that $\lim\limits_{(x,y)\to(0,0)}{x.y^2=0}$

Question

My try is:

$\left|x.y^2-0\right|<\left|\sqrt{x^2+y^2}.(x^2+y^2)\right|< \delta.\delta^2=\epsilon$

where $0<\sqrt{x^2+y^2}< \delta$ what did I miss here?

Answer 1

Writing $f(x,y) = x\cdot y^2$. Let $\varepsilon > 0$, setting $\delta = \varepsilon^{1/3}$, you have that $\sqrt{x^2 + y^2} \leq \delta \implies |f(x,y)- 0| \leq (\sqrt{x^2 + y^2})(\sqrt{x^2 + y^2})^2 \leq \varepsilon^{1/3} \cdot \varepsilon^{2/3} = \varepsilon$.

So, you have proven by constructing it that $\exists \delta, \sqrt{x^2 + y^2} \leq \delta \implies |f(x,y)- 0| \leq \varepsilon$.

Answer 2

Consider $x^2+y^2 <1$,

then $|y| \le 1$, and $y^2 \le |y|$.

Hence:

$|xy^2| \le |xy| \le (1/2)(x^2+y^2)$ .

Let $\epsilon >0$ be given.

Choose $\delta = \min (1,2\epsilon)$

Then $|x^2+y^2| \lt \delta$ implies

$|xy^2| \le$

$(1/2)(x^2+y^2) \lt (1/2)\delta =\epsilon$.

Used: $a^2+b^2 \ge 2|ab|$