# Homework Help: L'hospital's rule

1. Feb 1, 2007

### glid02

I have the equation A
44*e^(kt/5)+49((1-e^(kt/5))/k)

and I'm supposed to evaluate as k-->0

I think I'm supposed to apply l'hospital's rule to the second part of the equation, which would give
49*((1-t/5*e^(kt/5))/1)
which as k-->0 is
49*(1-t/5)

so the whole thing as k-->0 is
44+49*(1-t/5)

This isn't right, and I also tried l'hosital's rule on the first part of A, which would give 44*t/5 and this isn't right either.

What am I doing wrong?

Thanks.

Here's the whole question, in case I'm not reading it right:
Find the limit of this velocity for a fixed time t_0 as the air resistance coefficient k goes to 0.

2. Feb 1, 2007

### arildno

You have, by L'Hopital's rule:
$$\lim_{k\to{0}}49\frac{1-e^{\frac{kt}{5}}}{k}=\lim_{k\to{0}}49\frac{-\frac{t}{5}e^{\frac{kt}{5}}}{1}=-\frac{49}{5}t$$

3. Feb 1, 2007

### glid02

Oh yeah, that 1 should've disappeared from what I found.

Thanks a lot.