I've followed along in Stewart's calculus (last example in sec 10.4) up to this point:
I understand the that the polar slope formula can be found from product rule with a rectangular-to-polar parameterization of points. And I also understand all the that, vertical's are when $\frac{dx}{d\theta}=0$ as long as $\frac{dy}{d\theta}\neq 0$, and horizontals are when $\frac{dy}{d\theta}=0$ as long as $\frac{dx}{d\theta}\neq 0$.
And so that makes $\theta=\frac{3\pi}{2}$ a special case of $\frac{0}{0}$ and so it can be figured out by l'Hopital's... but I really don't get what he's doing. There might be some trig identity I'm missing, but where does that $-\frac{1}{3}$ come from and why set up l'Hopital's rule that way?
Previously, he had the slope $\frac{dy}{dx}=\frac{\cos\theta\left(1+2\sin\theta\right)}{\left(1-2\sin\theta\right)\left(1+\sin\theta\right)}$ but then he starts rearranging things without explanation.
$\lim fg =\lim f \cdot \lim g$ if both of these exist. Because $$\frac{1+2\sin\theta}{1-2\sin\theta} \to \frac{1+2(-1)}{1-2(-1)} =-\frac 13$$, it can be separated out from the initial limit. Then it remains to evaluate the limit of $$\frac{\cos\theta}{1+\sin\theta} $$ This is of the form $\frac 00$ and so LH can be applied here: $$\frac{-\sin \theta}{\cos \theta} \to \infty$$ as the numerator $\to -(-1) =1$ and the denominator $\to 0$.