Delta epsilon multivariable polar coordinates: Is $\lim_{r \to 0^+} (r \cdot f(\theta))$ always zero trivially?

Question

Suppose we want to prove the following using polar coordinates: $$\lim_{(x,y) \to (0,0)} \frac{xy}{\sqrt{x^2 + y^2}}$$

Subbing in as needed, I'd get:

$$\lim_{r \to 0^+}\frac{r \cos \theta \cdot r \sin \theta}{\sqrt{r^2 (\sin^2{\theta} + \cos^2{\theta})}}$$

and then after simplifying:

$$\lim_{r \to 0^+} r \cos{\theta} \sin{\theta}$$

From examples I've seen, others said that since $r$ goes to zero, it will take $cos$ and $sin$ with it to zero and therefore holds. Is this rigorous though? For example, would I need to demonstrate this:

$$r (\sin{\theta} \cos{\theta}) \leq r(1)$$

I ask because I'm unsure if there's a case where stating that

$$\lim_{r \to 0^+} (r \cdot f(\theta))$$

will ever cause problems, or that there is not any such function $f(\theta)$ that could ever make the above limit 'not zero' (assuming $r$ is not part of the function $f$).

---

Side question: I don't understand why $r$ is approached from the positive side only, I have just seen these in examples. Why not from the negative side?

Answer 1

It is important that sine and cosine are bounded functions of theta. So regardless of what $\theta$ is, $\sin\theta,\cos\theta$ are always between $-1$ and $1$ so that $$|r\cos\theta\sin\theta|\leq |r|$$ for ANY value of theta. That's why we are allowed to ignore them and just work with the $r$.

Answer 2

If you want to be a bit more rigorous in showing the final limit, note that $\vert \sin\theta \vert, \vert \cos\theta \vert \leq 1$, hence if $\epsilon > 0$ and if $0 < r < \epsilon$, we have $$\vert r \cos\theta\sin\theta \vert \leq \vert r \vert < \epsilon.$$