Is there a difference between instantaneous speed and the magnitude of instantaneous velocity?

Consider a particle that moves around the coordinate grid. After $t$ seconds, it has the position
$$
S(t)=(cos t, sin t) quad 0 leq t leq pi/2 , .
$$
The particle traces a quarter arc of length $pi/2$ around the unit circle. This means that the average speed of the particle is
$$
frac{text{distance travelled along the arc of the circle}}{text{time}}=frac{pi/2}{pi/2} = 1 , .
$$
However, since the motion of the particle is circular, the distance travelled is not the same as the displacement. The displacement of the particle would be $sqrt{2}$, and so the average velocity would be
$$
frac{text{straight line distance from initial position}}{text{time}} = frac{sqrt{2}}{pi/2} = frac{2sqrt{2}}{pi} text{ at angle of $frac{3}{4}pi$ with the positive $x$-axis} , .
$$
Here is the part I don’t quite understand: over an interval, the average speed of the particle is different from the magnitude of its velocity. In the above example, the former is $1$, whereas the latter is $frac{2sqrt{2}}{pi}$. However, the magnitude of the instantaneous velocity of the particle is the same as the instantaneous speed: here, they are both equal to $1$. We can mathematically prove this by considering the following limit
$$
|S'(t)| = lim_{h to 0}frac{|S(t+h)-S(t)|}{|h|}=lim_{h to 0}frac{sqrt{left(sin(t+h)-sin t right)^2+left( cos(t+h)-cos tright)^2}}{|h|} , ,
$$
which turns out to be equal to $1$. Hence, the magnitude of the instantaneous velocity is $1$. And clearly, the instantaneous speed of the particle is
$$
lim_{h to 0}frac{h}{h} = 1 , ,
$$
since the distance travelled along the arc between $S(t+h)$ and $S(t)$ is simply $h$ units. However, will this always be the case? Is the magnitude of the instantaneous velocity of a particle always equal to its instantaneous speed?