L'hospitals rulee major help

  • Thread starter angel_eyez
  • Start date
  • #1
19
0
l'hospitals rulee!!! major help

1. i don't know how to do this question
limx->infinity (x/x+1)^x. i keep gettin the rong answer, some heelp plzz:confused:




2.the right answer is 1/e but im not getting that, i tried many diff ways, i cant post my work it here because its too long and confusing



The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
529
1
Hint

Consider the series expansion for e^[1/x], when x is large.
 
  • #3
19
0
umm so when x is large 1/x = 0 so that means e^o = 1, im still lossttt
 
  • #4
529
1
Do you know the Taylor series for e^y, when y is small?
 
  • #5
19
0
noo..havent learned it =S
 
  • #6
19
0
i took the ln of both sides, and then found the derivative, and i got the answer as zero.. so i got ln y= 0
den i used the fact that e^lny = y...there fore y= e^0 =S..
 
  • #7
529
1
Hmmm, I'm not sure how to do it without recourse to a Taylor series expansion. It's easy to prove though that

e^x=1+x+x^2/2!+x^3/3+....

Have you done Taylor series at all?
 
  • #8
19
0
noo i haven't :|
 
  • #9
19
0
kk thnnxx for your help newayzz =)
 
  • #10
529
1
You suggested that L'hopital's rule should be used in the derivation. I played around with that and couldn't see it, but maybe someone else can work it out.

Let me know when you get the answer.
 
  • #11
Dick
Science Advisor
Homework Helper
26,260
619
You should have gotten something like x*ln(x/(x+1)) when you took the log. This is an infinity*0 form. To use L'Hopital you want to write it as 0/0 (or infinity/infinity). So write it as ln(x/(x+1))/(1/x). Now take derivative of the numerator and denominator and send x->infinity.
 
  • #12
dextercioby
Science Advisor
Homework Helper
Insights Author
13,023
576
There's no need for l'Hospital's rule, as this limit is elementary

[tex] \lim_{x\rightarrow \infty}\left(1-\frac{1}{x+1}\right)^{(x+1)\frac{x}{x+1}} =e^{-\lim_{x\rightarrow \infty}\frac{x}{x+1}}=e^{-1} [/tex]
 
  • #13
529
1
Dick, Brilliant!
 
  • #14
529
1
There's no need for l'Hospital's rule, as this limit is elementary

[tex] \lim_{x\rightarrow \infty}\left(1-\frac{1}{x+1}\right)^{(x+1)\frac{x}{x+1}} =e^{-\lim_{x\rightarrow \infty}\frac{x}{x+1}}=e^{-1} [/tex]
That's supposing that you know the Taylor series expansion of e^x. It's nice that you can prove it a completely different way.
 
  • #15
Dick
Science Advisor
Homework Helper
26,260
619
Dick, Brilliant!
Thanks. Wish I could say I invented it, but it's a pretty routine thing to do.
 
  • #16
dextercioby
Science Advisor
Homework Helper
Insights Author
13,023
576
That's supposing that you know the Taylor series expansion of e^x. It's nice that you can prove it a completely different way.
Pardon me ? I just the definition of "e". What's more elementary than that ??:rolleyes:
 

Related Threads on L'hospitals rulee major help

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