$\varepsilon-\delta$ proof of this multivariable limit

Question

We've to prove that $$ \lim_{(x,y)\to(0,0)} \frac{x^3+y^4}{x^2+y^2} =0 $$

Kindly check if my proof below is correct.

Proof

We need to show there exists $\delta>0$ for an $\varepsilon>0$ such that $$ \left| \frac{x^3+y^4}{x^2+y^2} \right| < \varepsilon \implies \sqrt{x^2+y^2}< \delta $$

Start $$ \left| \frac{x^3}{x^2+y^2} \right| <\left| \frac{x^3+y^4}{x^2+y^2} \right| < \varepsilon $$

Note $$ \left| \frac{x^3}{x^2+y^2} \right|= \frac{x^2|x|}{x^2+y^2}>\frac{(x^2-y^2)|x|}{x^2+y^2} $$

Therefore $$ \frac{x^2-y^2}{x^2+y^2}|x|<\varepsilon \tag{1} $$

Note $$ \frac{x^2-y^2}{x^2+y^2}|x|<|x|=\sqrt{x^2}<\sqrt{x^2+y^2}<\delta $$

So $$ \frac{x^2-y^2}{x^2+y^2}|x|<\delta \tag{2} $$

From $(1)$ and $(2)$, we can say $$ \delta=\varepsilon $$

Answer 1

That is not correct. You are supposed to find, for each $\varepsilon>0$, some $\delta>0$ such that$$\sqrt{x^2+y^2}<\delta\implies\left|\frac{x^3+y^4}{x^2+y^2}\right|<\varepsilon,$$and that's not what you did.

Note that$$\left|\frac{x^3}{x^2+y^2}\right|=|x|\frac{x^2}{x^2+y^2}\leqslant|x|$$and that$$\left|\frac{y^4}{x^2+y^2}\right|=y^2\frac{y^2}{x^2+y^2}\leqslant y^2\leqslant|y|$$if $y\in[-1,1]$. So, take $\delta=\min\left\{1,\frac\varepsilon2\right\}$. Then, if $\sqrt{x^2+y^2}<\delta$, then $y\in[-1,1]$ and $|x|,|y|<\delta\leqslant\frac\varepsilon2$, and therefore$$\left|\frac{x^3+y^2}{x^2+y^2}\right|\leqslant|x|+|y|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon.$$

Answer 2

There is a simple strategy for dealing with such limits, which is just switching to polar coordinates.
By setting $x=\rho\cos\theta, y=\rho\sin\theta$ we have $x^2+y^2=\rho^2$ and

$$ \left|\frac{x^3+y^4}{x^2+y^2}\right| = \rho\left|\cos^3\theta+\rho \sin^4\theta\right|. $$ The RHS is trivially bounded by $\rho(1+\rho)$, which is $\leq 2\rho$ if $\rho\leq 1$.
It follows that as long as $\rho\leq\frac{\varepsilon}{2}\leq 1$ we have $|f(x,y)|\leq \varepsilon$.

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If you want a sharp bound you may consider that for any sufficiently small $\rho\in\mathbb{R}^+$ the maximum of $|\cos^3\theta+\rho\sin^4\theta|$ is indeed $1$, so $\rho\leq\varepsilon$ is sufficient to ensure $|f(x,y)|\leq\varepsilon$.