Multivariable delta-epsilon limit proofs

Question

The Question asks: Find the delta epsilon limit if it exist $$ \lim \limits_{(x, y) \to (0, 0)} \ \frac{x^2 y^3}{2x^2 + y^2} = 1$$

What I've done so far:

$$ \mid {\frac{x^2 y^3}{2x^2 + y^2} - 1 \mid} < \epsilon \quad \quad \quad \quad0 < \sqrt{x^2 + y^2} < \delta $$

Now, since $x ^2 \leq 2x^2 + y^ \implies \frac{x^2}{2x^2 + y^2} \leq 1$ Then: $$ \frac{x^2 \mid y^3 \mid}{2x^2 + y^2} \leq \mid y^3 \mid = \sqrt{y^2}^3 = \quad ...$$

Here is where I am stuck, I can't make $\sqrt{y^2}^3$ into $\sqrt{ x^2 + y^2} $.

How should I proceed further?

Using Stewart's Calculus (8th ed) 14.2 Limit proof method

Answer 1

For both $x,y\ne 0$, then \begin{align*} \dfrac{|x|^{2}|y|^{3}}{2x^{2}+y^{2}}&\leq\dfrac{|x|^{2}|y|^{3}}{2\sqrt{2}|x||y|}\\ &=\dfrac{1}{2\sqrt{2}}|x||y|^{2}\\ &\leq\dfrac{1}{2\sqrt{2}}\sqrt{|x|^{2}+|y|^{2}}(|x|^{2}+|y|^{2}), \end{align*} if either $x=0$ or $y=0$ (but not both), the above inequality is still true. So \begin{align*} \dfrac{|x|^{2}|y|^{3}}{2x^{2}+y^{2}}\leq\dfrac{1}{2\sqrt{2}}\sqrt{|x|^{2}+|y|^{2}}(|x|^{2}+|y|^{2}),~~~~(x,y)\ne(0,0). \end{align*} Given $\epsilon>0$, for all $0<\sqrt{|x|^{2}+|y|^{2}}<2^{1/2}\epsilon^{1/3}$, then $\dfrac{|x|^{2}|y|^{3}}{2x^{2}+y^{2}}\leq\dfrac{1}{2\sqrt{2}}\sqrt{|x|^{2}+|y|^{2}}(|x|^{2}+|y|^{2})=\dfrac{1}{2\sqrt{2}}(|x|^{2}+|y|^{2})^{3/2}<\epsilon$.

The limit is zero.

Another way: \begin{align*} \dfrac{x^{2}}{2x^{2}+y^{2}}\leq 1,~~~~(x,y)\ne(0,0), \end{align*} so \begin{align*} \dfrac{x^{2}|y|^{3}}{2x^{2}+y^{2}}\leq|y|^{3}=(y^{2})^{3/2}\leq(x^{2}+y^{2})^{3/2}. \end{align*}

Answer 2

In fact

$$\left|\frac{x^2y^3}{2x^2+y^2}\right| \leq \frac{1}{2}|y^3|$$ which is equivalent to

$$2x^2\leq 2x^2+y^2.$$