Suppose I have a parametric equation $x(t)=t^3$, $y(t)=\sin(t)$.
Then, $$\frac{dy}{dx}=\frac{\cos(t)}{3t^2}$$ At $t=0$ derivative doesn't exist. So, let's take a limit: $\lim_{t\to 0}\frac{\cos(t)}{3t^2}=\infty$. So, there is a vertical tangent line at $t=0$.
Now, suppose I have a parametric equation $x(t)=t^3$, $y(t)=\cos(t)$.
Then, $$\frac{dy}{dx}=\frac{-\sin(t)}{3t^2}$$ At $t=0$ derivative doesn't exist. So, let's take a limit: $\lim_{t\to 0^+}\frac{-\sin(t)}{3t^2}=-\infty$, $\lim_{t\to 0^-}\frac{-\sin(t)}{3t^2}=\infty$. Since the limit doesn't exist, then there is no vertical tangent line at $t=0$.
Now, consider the polar curve $r(\theta)=\cos(\theta)$. Then, $$\frac{dy}{dx}=-\cot(2\theta)$$
Clearly, the limit doesn't exist at $\theta=0$, but tangent line exists and it equals $x=1$.
I thought that if limit exists and finite, then this is the "normal" case, if limit exists and infinite, then there is a vertical tangent line, if it doesn't exist, then there is no tangent line.
Where am I wrong?
Consider the curve parameterized by $x(t) = \sin(t),$ $y(t) = t^3.$ For this curve, $\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{3t^2}{\cos(t)},$ and
$$ \lim_{t\to 0} \frac{3t^2}{\cos(t)} = 0, $$
so by the rules you've given, you have a horizontal tangent at $t=0.$
But the curve is an exact image of the one parameterized by $x(t) = t^3,$ $y(t) = \sin(t)$ under reflection through the line $y = x.$ In parametric curves like these, we care about the geometric shape, not whether the curve can be described by $y$ as a function of $x.$ A horizontal tangent at $t=0$ on one curve ought to correspond to a vertical tangent on the other.
Another way to look at the curve $x(t) = t^3,$ $y(t) = \sin(t)$ is:
$$ \lim_{t\to 0} \frac{\mathrm dx}{\mathrm dy} = \lim_{t\to 0} \frac{3t^2}{\cos(t)}= 0. $$
But note that this rule cannot distinguish when a curve has a cusp. If you're concerned about that, you might want to use tangent vectors rather than tangent lines.