If friction serves as the centripetal force then can it be greater than the centripetal force itself? I know the title looks vague and the concept also looked strange to me when I read this.

In the image, it is pretty clearly written that to avoid skiding friction force serves as the centripetal force. But just a line below, the equation says :
$$Centripetal\; \;force≤Force \; \; of \; \;friction $$
I don't know if it is true but I don't think it is true. If friction is the only horizontal force here then why should it be smaller than centripetal force ?
Correct me if I am wrong somewhere.
 A: I think what the book means is "centripetal force $\leq$ maximum force of friction".
Think of the simple case of a block on a slightly inclined plane: the maximum force that friction can exert is $\mu F_\perp$, where $F_\perp$ is the component of the block's weight that is perpendicular to the plane; if this maximum force is bigger than the parallel component of the weight (which is trying to make the block slide), then friction will only exert a force equal to the parallel weight, and no more, otherwise the block would start climbing upwards.
So what I think your book means is that, in order not to have skidding, the centripetal force required should be smaller than the maximum force of friction $\mu F_\perp$.
A: Centripetal force is the name given to "a force that acts on a body moving in a circular path and is directed toward the center around which the body is moving". There is no "basic" centripetal force; it is just the name given to a true force- such as gravity or friction- that acts as described in the first sentence.  As other answers state, the force of friction must not be exceeded for that force to be a centripetal force.
(A centrifugal force is a fictitious force that appears in a non-inertial rotating coordinate system.)
A: Well, I Don't think it's an error, it is just terribly phrased.
It's more like the maximum centripetal force on the object will be the maximum value of static friction.
For angular velocities lower than the max, friction will adjust accordingly.
A: Static friction will adjust itself (up to a maximum) to whatever value is required to prevent slipping.
A: Talking about skidding. You can take circular turning into consideration where you are driving  your car on circular road. Friction acts towards the centre and so does , centripetal acceleration.
Centripetal force = $\frac{mv^2}{r}$. I hope you know this. If not , then just comment.
Friction always acts in the direction opposite to direction where it can slide. For this situation  , you can say that your frictional force = $\frac{mv^2}{r}$. Since Friction force = m*a where a = $\frac{v^2}{r}$.
This force is also equal to μN = μMg.
So , it is just that they have written the same thing.
The book states that frictional force $\ge$ centripetal force. You have to make sure for is static or limiting friction. Centripetal force should be less than or equal to this. Otherwise , kinetic friction may occur and then , sliding may happen.
So , I hope you know now what happens if centripetal force is greater than the force itself or frictional force.
Do let me know if you have any difficulty   :)
A: Friction is reactive force, it is always same as a force that is oposing it. There is of course maximum frictional force and centrifugal force can exceed this so then you have some sliding. Since it has a role of centripetal force in this example these are same forces so to me too this seems strange. That equation up there would make more sense if it compared friction with centrifugal force with condition being that centrifugal must be less than maximal friction.
