I'm trying to construct an $\varepsilon-\delta$ proof for the following limit, $$\lim_{(x, y) \rightarrow (0, 0)}\frac{4xy}{x^2 + y^2 + 2} = 0,$$
and I would like to see if my proof is sound having converted cartesian coordinates to polar coordinates.
My proof is as follows:
Let $x = r\cos\theta$, $y = r\sin\theta$ with $r = \sqrt{x^2 + y^2}$, then
$$\lim_{(x, y) \rightarrow (0, 0)}\frac{4xy}{x^2 + y^2 + 2} = \lim_{r \rightarrow 0}\frac{4r^2\cos\theta\sin\theta}{r^2+2}$$.
We note that $\forall\theta\in\mathbb{R}, 0 \leq|\cos\theta\sin\theta| < 1$.
Now, suppose that $0 < |r| < \delta$, for $\delta > 0$. We take $\delta < \frac{\varepsilon}{4}$, with $\varepsilon > 0$, then $0 < |r| < \delta < \frac{\varepsilon}{4}$.
$$
\begin{align*}
\left|\frac{4r^2\cos\theta\sin\theta}{r^2+2} – 0\right| &< \left|\frac{4r^2\cos\theta\sin\theta}{r^2}\right|\\ &= 4|\cos\theta\sin\theta| \\
&<4|r||\cos\theta\sin\theta|\\
&<4|r| \\
&<4 \cdot \frac{\varepsilon}{4} \\
&= \varepsilon
\end{align*}$$
Thus for $0<|r|<\delta$, we have $\left|\frac{4r^2\cos\theta\sin\theta}{r^2+2}\right| < \varepsilon$. Hence, we conclude that
$$\lim_{(x, y) \rightarrow (0, 0)}\frac{4xy}{x^2 + y^2 + 2} = \lim_{r \rightarrow 0}\frac{4r^2\cos\theta\sin\theta}{r^2+2}$$
exists and is equal to $0$.
Anyways, any feedback of how I could improve the proof (if it is correct) or correct anything would be greatly appreciated.
Best Answer
Your argument is wrong, because you made an $|r|$ appear out of thin air. Without it, your expression does not go to zero.
Using polar coordinates here is overkill. What you should note is that the denominator is away from zero, so it is irrelevant (and the right bound to use was $2$ and not $r^2$). Then you can do the following: $$ \Big|\frac{4xy}{x^2+y^2+2}\Big|\leq\frac{4|xy|}{2}=2|xy|\leq x^2+y^2. $$ Now you can take $\delta=\varepsilon^{1/2}$.