Is motion smooth? It's obvious that for every particle velocity is smooth i.e it cannot undergo sudden finite change in its position in infinitisiminal time.
Similarly any particle's velocity cannot undergo a change instantaneously (Infinite acceleration can't happen, intuitively).
Does this pattern apply to higher time derivatives of position like jerk? If yes then till how much higher derivative? 10th? 100? Infinite?
 A: Typically these higher derivatives are assumed to be smooth.
The key question will be what causes a discontinuity in the n-th derivative.  If you focus on classical mechanics, the forces on an object boil down to the positions of particles in the system, which are continuous.  This means there would need to be a discontinuity (such as a divide by zero) in the equations of motion in order to have a non-smooth higher derivative.
When you push towards quantum mechanics, the terms get really murky quickly, because position ceases to be a single observable number.  But if you stick to classical mechanics, we find things stay nice and smooth.
Now that being said, when modeling real systems, we very often assume instantaneous changes in velocity or higher derivatives.  This is because, in many cases, we can get away with ignoring the precise acceleration or jerk function and treat it as-if it were a simple discontinuous system.  A straight forward example of this is a billiard ball collision.  For most intents and purposes, this collision is "instantaneous" and the velocity of the balls changes in a discontinuous manner.  However, if we look closer, with a slow motion camera, we find that the collision is not actually instantaneous -- position and its derivatives smoothly change over time.  In fact, if you look hard enough, you can even see the ripple as the effects of the impact race across the surface of the ball.  But, for the purposes of determining the result of a trick shot, these fine details are immaterial, and calculations assuming an instantaneous change in velocity are used.
A: In order for jerk to be non-zero, the acceleration $a=\frac{\mathbf{d}v}{\mathbf{dt}}$ must be time-dependent:
$$a=\frac{\mathbf{d}v}{\mathbf{dt}}=f(t)\tag{1}$$
That's because the derivative of any number, no matter how large, is always zero.
But if $(1)$ applies the jerk becomes:
$$j=\frac{\mathbf{d}a}{\mathbf{dt}}=f'(t) \neq 0$$
If we take a very simple case, where:
$$a=a_0+a_1t$$
then:
$$j=(a_0+a_1t)'=a_1\tag{2}$$
And the even higher derivatives all become zero.
$(2)$ also shows that $a$ could be very large (large $a_0$) and yet $j$ might be very small (small $a_1$)
