L'hospital's rule

  • Thread starter glid02
  • Start date
  • #1
54
0
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.
 

Answers and Replies

  • #2
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
132
You have, by L'Hopital's rule:
[tex]\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[/tex]
 
  • #3
54
0
Oh yeah, that 1 should've disappeared from what I found.

Thanks a lot.
 

Related Threads on L'hospital's rule

  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
9
Views
5K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
21
Views
4K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
12
Views
1K
Top