Horizontal tangent line of a parametric curve Suppose $x=t^2,y=t^3$ is a parametric curve. Here's a quote from my textbook:

The origin, which corresponds to $t=0$, is a singular point of the parametric curve, because $dx/dt=2t,dy/dt=3t^2$ are both zero when $t=0$.

So far so good.
But then they write:

However, the curve has a horizontal tangent line at the origin, because for all $t\neq 0$:
  $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3}{2}t$$
  And thus:
  $$\lim_{t\to 0^+} \frac{dy}{dx}=\lim_{t\to 0^-} \frac{dy}{dx}=0$$

It looked a little odd for me. Nevertheless, I decided to use the same argument to show that the parametric curve $x=2\cos t - \cos (2t), y=2\sin t - \sin(2t)$ has a horizontal tangent line at $t=0$, that is at $(1,0)$.
However my professor said that this is wrong ("because the derivative is not zero" - indeed, $\frac{dy}{dx}\Big|_{t=0}$ is undefined - "$0/0$").
So who is right? Is the existence of the limit a sufficient condition for the (horizontal) tangent line to exist, as my textbook says, or not? I'm confused.
Thanks.
Short version of the question: can a parametric curve have a horizontal tangent line at a singular point? 
I.e. is $\lim_{t\to 0} \frac{dy}{dx}=0$ a sufficient condition for a horizontal line to exist (at $t=0$)? (even if the derivative $\frac{dy}{dx}\Big |_{t=0}$ itself doesn't exist).
 A: Notice that both your curves are algebraic curves, of equations $x^3 - y^2 = 0$ and $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ respectively.
We say that the line $ax + by + c = 0$ is tangent at the point $(x_0, y_0)$ to the curve given by $F(x,y) = 0$ if and only if $(x_0, y_0)$ is a multiple root of the system $\begin{cases} F(x,y) = 0 \\ ax + by + c = 0 \end{cases}$ (this is the definition that was originally used by algebraic geometers).
In your case, the points are $(0,0)$ in the first case and $(1, 0)$ in the second, and the line to be checked is $y=0$.
Plugging $y = 0$ into $x^3 - y^2 = 0$ gives $x^3 = 0$, which indeed has $x=0$ as a triple root, therefore $y=0$ is tangent to $x^3 - y^2 = 0$ at $(0,0)$.
Plugging $y = 0$ into $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ gives $(x^2-1)^2 - 4(x-1)^2 = 0$, or equivalently $(x-1)^2 ((x+1)^2 - 4) = 0$, or again $(x-1)^3 (x+3) = 0$, which indeed has $x=1$ as a triple root (the root $-3$ being simple, meaning that at the point $(-3, 0)$ the line $y=0$ intersects the curve as a secant), therefore $y=0$ is tangent to $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ at $(1,0)$.

Since the bounty giver is still unsure, let us follow Wikipedia and compute the intersection number in the first case. Wikipedia gives several methods, one of them is to consider the ideal generated by $y^2 - x^3$ and by $y$ in $\Bbb R[[x,y]]$ (the ring of formal series in $x$ and $y$) and compute the dimension of the quotient vector space $\Bbb R [[x,y]] / (x^3-y^2, y)$.
In the quotient space, both $y$ and $x^3 - y^2$ will become $0$, which implies that $x^3$ also becomes $0$. Therefore, the only powers of $x$ and $y$ that survive in the quotient set are $x^0 = y^0 = 1$, and $x$ and $x^2$ - a total of $3$ linearly independent powers, so $3$ will be the intersection number of $y=0$ and $x^3-y^2$ at $(0,0)$. Since everything $\ge 2$ means tangency, this shows that the line $y=0$ is tangent to $x^3-y^2$ at $(0,0)$.
The same thing could be done with what Wikipedia denotes by $I_{(0,0)}$, letting $P = y$ and $Q = x^3 - y^2$. Applying the properties that you see on that page (the numbers above equal signs are the Wikipedia properties that I apply),
$$I_{(0,0)} (y, x^3-y^2) \overset 6 = I_{(0,0)} (y, x^3) \overset 5 = I_{(0,0)}(x,y) + I_{(0,0)}(x,y) + I_{(0,0)}(x,y) \overset 4 = 1 + 1 + 1 = 3 .$$

The same computations could be done for the second example, it is just that they are more tedious. First, let us translate the curve such that the cusp moves from $(1,0)$ to $(0,0)$ (because, for simplicity, Wikipedia's formulae are given only for $(0,0)$). To do this we shall make the change of variable $x = u + 1$, which will lead to $(u^2 +2u +y^2)^2 -4(u^2+y^2)$. Next,
$$I_{(0,0)} ((u^2 +2u +y^2)^2 -4(u^2+y^2), y) \overset 6 = I_{(0,0)} (u^4 + 4u^3, y) = I_{(0,0)} (u^3 (u+4), y) \overset 5 = I_{(0,0)} (u^3, y) + \\ I_{(0,0)} (u+4, y) = I_{(0,0)} (u, y) + I_{(0,0)} (u, y) + I_{(0,0)} (u, y) + I_{(0,0)} (u+4, y) \overset {3, 4} = 1+1+1+0 = 3 .$$
Again, we obtain an intersection number $\ge 2$, which means tangency.

What others have correctly stated is that the two plane curves given in the problem are not differentiable submanifolds of $\Bbb R^2$ at the singular points under discussion, therefore they do not fit into the framework of diferential geometry, therefore we may not speak of their tangent spaces at the singular points as defined in differential geometry (a proof that an almost identical curve cannot be a smooth submanifold can be found here). Counterintuitively, though, they do have tangent lines! There is no contradiction, the thing is that we use two meanings for "being tangent" that are not synonymous: one comes from differential geometry ("tangent space"), and it doesn't apply here, the other from algebraic geometry ("intersection number"), and it does apply here. In the latter framework, we may speak about "a given line being tangent (or not) to a curve in a point" without speaking of "the tangent space at that curve in that point".
$y=0$ is tangent to those curves at the specified points. Those curves do not have nicely defined tangent spaces at those points. 
A: In order to find the tangent at some special point you should not try to compute the limit of $y'$ but the actual limit of secant directions at that point. Therefore in your first example you get
$$m_+:=\lim_{t\to0+}{y(t)-y(0)\over x(t)-x(0)}=\lim_{t\to0+}{t^3\over t^2}=0\ ,$$
and similarly $m_-=0$. In your second example you have
$${y(t)-y(0)\over x(t)-x(0)}={2\sin t-\sin(2t)-0\over 2\cos t-\cos(2t)-1}={2\sin t(1-\cos t)\over2\cos t(1-\cos t)}=\tan t\qquad(t\ne0)\ ,$$
and therefore
$$m_+=m_-=\lim_{t\to0}{y(t)-y(0)\over x(t)-x(0)}=0\ .$$
A: the formula $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$ is valid for computing the derivative at points where $\frac{dx}{dt}\neq 0$. At singular points, i.e. points at which $$\frac{dy}{dt}=\frac{dx}{dt}=0$$ the above formula is not valid. However, this does not imply that the derivative does not exist; only that it is not computable by the above formula. 
In such cases, a straightforward application of the definition of derivative (as already done by user Christian Blatter in his answer) can provide the answer:
$$
\frac{dy}{dx}\bigg|_{(0,0)}=\lim_{t\to 0^+}\frac{y(t)-y(0)}{x(t)-x(0)}=\lim_{t\to 0^-}\frac{y(t)-y(0)}{x(t)-x(0)}=\lim_{x\to0^\pm}\frac{t^3}{t^2}=0
$$
which shows that the derivative exists at $(0,0)$ and so does the tangent line at $(0,0)$, which is no other than the horizontal axis itself. 
P.S.1: Note that, when confined to each branch seperately:
$$
y(x_0)=y(x(0))=y(0)=0, \  \ x_0=x(0)=0 \\ y(x)=y(x(t))=y(t)=t^3, \ \ x=x(t)=t^2
$$ 
So, on each one of the functions $y(x)=x^{3/2}$, $y(x)=−x^{3/2}$ i.e. on each one of the branches of the graph below, the following rates of change are identified:
$$\frac{y\big(x(t)\big)−y(x_0)}{x(t)−x_0}=\frac{y(t)−y(0)}{x(t)−x(0)}$$
On the other hand, since $x=t^2\neq 0$ when $t\neq 0$ it should be clear that
$$t\to 0\Leftrightarrow x\to 0$$
At this point, it remains to invoke a classical proposition on the limit of a composite function, claiming that:

Let the real functions $f,g$, for which $lim_{x\rightarrow x_0}g(x)=g_0\in \mathbb{R}$ and $lim_{g\rightarrow g_0}f(g)=l\in \mathbb{R}$.
  If, furthermore $g(x)\neq g_0$ "close to  $x_0$" (i.e. in some interval $(a,x_0))$, then 
  $$
\lim_{x\rightarrow x_0}f\big(g(x)\big)=\lim_{g\rightarrow g_0}f(g)=l
$$

Seting $f(x):=\frac{y(x)−y(0)}{x−0}$ and $g(x):=x(t)$ and applying the proposition we get that:
$$\lim_{t\to 0}\frac{y(t)−y(0)}{x(t)−x(0)}\equiv \lim_{t\to 0}\frac{y\big(x(t)\big)−y(0)}{x(t)−x(0)}=\lim_{x\to 0}\frac{y(x)−y(0)}{x−0}=\frac{dy}{dx}\bigg|_0$$
(where the far left term is computed using side limits: for the upper branch  $t>0$ thus $t\to0\Leftrightarrow t\to0^+$ while for the lower branch $t<0$ thus $t\to0\Leftrightarrow t\to0^-$).
P.S.2: Regarding your final question: I do not believe that the existence of the limit is a sufficient condition, because generally, a differentiable function at $x_0$ need not have a continuous derivative at $x_0$: In other words, there may well be a situation at which: $\lim_{x\to x_0}f'(x)\neq f'(x_0)$. 
However, this is not the case for the semicubical parabola, as we can see from its graph 

which is probably what the author of your textbook had in his mind. He computes the limit of the derivative just as if we can imagine the tangent "slipping" accross either of the branches to "continuously" become horizontal. In my opinion, he implicitly uses the assumption that the derivative of the semicubical parabola is a continuous function; and he probably does that, based on the shape of the graph.   
A: You are referring to two different things here..
For the semi-cubical parabola curve $y=x^{3/2}$ there is horizontal tangent at the cusp y=0 or t=0.
For cardoid, depending on the cusp contact point chosen, slope of cusp tangent varies, look at all cusps of  epicycloids:
EDIT 1:
Since it is passing through infinite curvature which is rate of change of tangent inclination, the slope is changing very fast and undefined.
It can be also seen at the point $(1,0)$ Changing the multiple angle
$$ x=2\cos t - \cos (n t), y=2\sin t - \sin( n t) $$ 
has the effect of influencing curvature. It changes from (positive for circle) to negative for looped case, via the present central cardoid case of infinite curvature.
The direction of tangent is vertical, indeterminate and vertical respectively

