newtonian mechanics - How can centripetal acceleration not change speed?

I have seen a question similar to this asked online, but the answers do not make complete sense to me. In studying centripetal acceleration, the textbook I am using argues that the centripetal acceleration must be perpendicular to the tangential velocity, otherwise the speed of the object would change. However, wouldn't the speed still change if the acceleration were perpendicular (see this post).

Now, I have seen the arguments online that this is just a misconception because the perpendicular acceleration is lasting for an infinitesimal (not finite) amount of time so it will not change the speed (significantly). However, if the centripetal acceleration is lasting for such a short period of time that it doesn't change the speed of the object, won't it also not change the direction of the object at all (since it is "so small")? Of course, you can integrate the infinitesimal accelerations to "eventually" change the direction, but in doing this aren't you also integrating the "negligible" errors? Am I misunderstanding something? Thank you!

Edit: I know that the question I linked to is very similar, however I do not think the answers to that question fully answer my question. The question linked to started out without acknowledging the infinitesimal duration of the centripetal acceleration. This is the main focus of my question.

Answer

The speed is the length of the velocity vector. An acceleration vector that has a positive component along the direction of the velocity will increase the length and thus the speed. Similarly, an acceleration vector with a negative component along the direction of the velocity will decrease the speed. If the acceleration vector is perpendicular to the velocity, it's component along the direction of the velocity vector is zero. Based on the two statements above, what effect would you expect that acceleration vector to have on the speed?

We can quantify the above by calculating the derivative of speed using the product rule (I'll do it in two-dimensions rather than three for simplicity). The speed is $$ s = |\vec v| = \sqrt{v_x^2+v_y^2}$$ where $v_x$ and $v_y$ are the components of the velocity vector. Using the chain rule and then the product rule and then the chain rule again gives $$\frac{ds}{dt} = \frac{1}{2\sqrt{v_x^2+v_y^2}} \left(2 v_x\frac{d v_x}{dt}+2 v_y\frac{d v_y}{dt}\right)\\=\frac{1}{s}\vec v\cdot \vec a $$ where we used the fact that the acceleration vector has components $$ \vec a = \left(\frac{dv_x}{dt},\frac{dv_y}{dt}\right).$$

From this equation we see that if the velocity and acceleration are perpendicular , $\vec v\cdot \vec a=0$ so the speed is not changing.