# L'Hospital's Rule question: Find \lim_{t\rightarrow\infty}v(t)?

## The downward velocity $v$ of a skydiver with nonlinear air resistance can be modeled by: $v = v \left(t\right) = - A + R A \left(\frac{{e}^{B t + C} - 1}{{e}^{B t + C} + 1}\right)$ Note please apply the rule if necessary, as it is in this section. this is Calculus I/Single Variable any further advanced explanations can be added if you want...

Nov 26, 2016

I found $A \left(R - 1\right)$ but check my maths because I did it in a hurry and I may have overlooked something...

#### Explanation:

Have a look:

Nov 26, 2016

Depends on $B$ sign:
$B > 0 \to v \left(\infty\right) = A \left(R - 1\right)$
$B < 0 \to v \left(\infty\right) = A \left(R \frac{{e}^{C} - 1}{{e}^{C} + 1} - 1\right)$

#### Explanation:

Supposing $B > 0$

$\frac{{e}^{B t + C} - 1}{{e}^{B t + C} + 1} = {e}^{B t} / {e}^{B t} \left(\frac{{e}^{C} - {e}^{- B t}}{{e}^{C} + {e}^{- B t}}\right) = \frac{{e}^{C} - {e}^{- B t}}{{e}^{C} + {e}^{- B t}}$

Now ${\lim}_{t \to \infty} {e}^{- B t} = 0$ so

$v \left(\infty\right) = {\lim}_{t \to \infty} \left(- A + R A \left(\frac{{e}^{C} - {e}^{- B t}}{{e}^{C} + {e}^{- B t}}\right)\right) = A \left(R - 1\right)$

Supposing $B < 0$

${\lim}_{t \to \infty} {e}^{B t} = 0$ so

$v \left(\infty\right) = A \left(R \frac{{e}^{C} - 1}{{e}^{C} + 1} - 1\right)$

l'Hopital's rule is not needed because the limit is well defined.