# Limits involving absolute values

1. Mar 26, 2015

This is actually a physics problem, but since my question is really about the math involved, I decided to post it in the calculus subforum.
I'm supposed to get from the term:
$$\lim_{\Delta t → 0} |\vec{v}_r (t + \Delta t)| \frac{\sin \Delta \theta}{\Delta t}$$
To:
$$v_r (t) \frac{d\theta}{dt}$$
$\theta$ is a function of $t$ (so $\Delta \theta$ approaches zero as $\Delta t$ approaches zero), and $v_r$ is the scalar component (not the magnitude) of the vector $\vec{v}_r$, in a particular direction.
Computing the limit as $\Delta t$ approaches zero:
$$\lim_{\Delta t → 0} |\vec{v}_r (t + \Delta t)| \frac{\sin \Delta \theta}{\Delta t} = |\vec{v}_r (t)| \frac{d\theta}{dt} = |v_r (t)| \frac{d\theta}{dt}$$
How do I get rid of those absolute value bars?

2. Mar 26, 2015

### Staff: Mentor

Then by this logic $\lim_{\Delta t}\frac{\Delta y}{\Delta t} = \frac 0 0$, assuming as you did that y is a function of t.

What they're probably doing here is using the idea that, if u is close to zero, then $\sin(u) \approx u$.
Is $v_r$ the magnitude of $\vec{v_r}$? If so $v_r$ would be a nonnegative scalar.

3. Mar 26, 2015

$v_r$ is not the magnitude of $\vec{v}_r$. It is the scalar component of $\vec{v}_r$ in a particular direction. It can be positive, negative, or zero.
Its absolute value is the magnitude of $\vec{v}_r$.