why the curvature of a spiral in its origin is not infinity? It can be shown that the curvature of a spiral $\bf{r}(\rm t)=t(\cos t, \sin t)$ is given by 
\begin{eqnarray}
    \kappa(t) = \frac{t^2 + 2}{(\sqrt{1+ t^2})^3}
\end{eqnarray}
Given that the radius at $t=0$ is $0$, I would think that the curvature is
infinity. Still $\lim_{t \to 0} \kappa(t) = 2$. 
 A: I think the problem here is that you're confusing polar radius with radius of curvature.
The polar radius is $0,$ but the radius of curvature isn't. The curve doesn't knot into a point at the origin. Also, from a plot of the curve, the osculating circle is far from being degenerate. Thus, the radius of curvature is positive and finite.
A: I finally answered my question with the help of WolframAlpha.
After a few tests here is what we have.
We have a user who removed his name from this blog, so I do not have who to give credits to. He suggested that the figure was not a spiral and plotted something like:
$t$ between -5, 5 ">
To get the picture in your browser just click
first figure
It seems as if the picture is not a spiral but it is. 
See what happens when we pick $t \in [0, 20]$.

You can load this figure in your browser by clicking
here
Now to understand what is going on let us choose $t \in [-1,10]$.
Here is the figure.

You can load the picture in your browser by clicking
here
It is now clear that for $t=0$ you can make a small circle with radius
$1/2$, center at $[0,1/2]$ tangential at the $x$ axis at $(0,0)$. 
The curvature of this circle is $\kappa=2$. The figure does not lie
but my intuition was wrong and I thought that this spiral collapse 
to a point $(0,0)$ while indeed it just passes through that point
tangentially. 
A: Here are some insights as to why the curve has the behaviour it has.
Symmetry around zero
The first thing to notice is that $t,\sin t$ are odd functions whereas $\cos t$ is even. Hence $t\cos t$ is odd and $t\sin t$ is even. This implies that the curve of the form $\mathbf r(t)=(odd,even)$ must be symmetrical around the $y$-axis. Due to this symmetry, the center of curvature for $t=0$ must lie on the $y$-axis. Let $(0,y)$ be the center of a circle through $(0,0)$ and $\mathbf r(t)$. Then we have:
$$
y^2=t^2\cos^2 t+(y-t\sin t)^2\\
\Updownarrow\\
y=\frac t{2\sin t}
$$
This clearly tends to $y=1/2$ as $t$ tends to zero. Thus we have placed the center of curvature for $t=0$ at $(0,y)=(0,1/2)$ which verifies $\kappa(0)=2$.
Resemblance to a parabola
First some useful facts to note:


*

*Tangents are first order differential approximations. 

*Circles of curvature coincide locally with second order differential approximations.


Now what happens near $t=0$? We have the first order approximations $\cos t\approx 1$ and $\sin t\approx t$ which allows us to derive the second order approximations:
$$
\mathbf r(t)=t(\cos t,\sin t)\approx t(1,t)=(t,t^2)
$$
and so the graph of $\mathbf r(t)$ resembles the graph of the quadratic function $f(t)=t^2$ near $t=0$.
General second order approximations
More generally, one can show that we have the second order approximation:
$$
\mathbf r(t)\approx t \mathbf e_1+t(t-t_0)\mathbf e_2
$$
near $t=t_0$ where $\mathbf e_1,\mathbf e_2$ are the perpendicular unit vectors:
$$
\begin{align}
\mathbf e_1&=(\cos t_0,\sin t_0)\\
\mathbf e_2&=(-\sin t_0,\cos t_0)
\end{align}
$$
In fact we have the parabola $f(t)=t(t-t_0)$ in the coordinate system defined by the basis $\mathbf e_1,\mathbf e_2$. This has zeros for $t=0$ and $t=t_0$ corresponding to the origin $(0,0)$ and the point $\mathbf r(t_0)$ on the curve. Only for $t_0=0$ we have a double root so that this parabola has its vertex at the curve point.
