Lim n[tex]\rightarrow[/tex][tex]\infty[/tex] (1+(x/n))[tex]^{n}[/tex]

  • #1
lim n[tex]\rightarrow[/tex][tex]\infty[/tex] (1+(x/n))[tex]^{n}[/tex]

I have no idea where to start. Can anyone who knows what to do give me a hint or tell me the first step? Thanks
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,847
967
That looks an awful lot like an exponential function to me. Take the logarithm of [itex](1+(x/n))^n[/itex] and you have a limit, as n goes to infinity, of the form [itex]0*\infty[/itex]. That can be rewritten as so that it is of the form "0/0" and, even though here n is an integer, you can use L'Hopital's rule.
 
Last edited by a moderator:
  • #3
That looks an awful lot like an exponential function to me. Take the logarithm of [itex]1+(x/n))^n and you have a limit, as n goes to infinity, of the form [itex]0*\infty[/itex]. That can be rewritten as so that it is of the form "0/0" and, even though here n is an integer, you can use L'Hopital's rule.

I'm pretty sure I can't pull the one out of the parenthesis like that. Like when you foil you always have a middle term? So with this we should have n+1 terms, but if we take the one out it chages the equation and we only have two terms if we expand it
 
  • #4
Maybe I'm wrong. If I can't come up with something better I will do that
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,847
967
The 1 was supposed to be in the parentheses! I have edited my previous post.

The logarithm of [itex](1+ (x/n))^n[/itex] is n log(1+ (x/n)) which, as I said, is of the form "[itex]0*\infty[/itex]". You can write that as log(1+ (x/n))/(1/n) so that it is now of the form "0/0". Apply L'Hopital's rule treating the n as a continuous variable.
 
  • #6
alright. thanks so much!
 

Related Threads on Lim n[tex]\rightarrow[/tex][tex]\infty[/tex] (1+(x/n))[tex]^{n}[/tex]

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
5
Views
4K
Replies
4
Views
6K
Replies
1
Views
1K
Replies
5
Views
5K
Replies
2
Views
1K
Replies
3
Views
2K
Top