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Lim as n->infinity help

  • Thread starter shan
  • Start date
57
0
It seems to an area I'm really fuzzy on. Anyway, the question I've having problems with is

lim as n->infinity[tex]
(1 + \frac{1}{n+1}) ^ {3-n}
[/tex]

At first, I thought I could evaluate the stuff inside the bracket first and then take care of the power ie

lim as n->infinity[tex]
(1 + \frac{1}{n+1})
[/tex] = 1

so lim as n->infinity[tex]
(1 + \frac{1}{n+1}) ^ {3-n} = 1 ^ {3-n} = 1
[/tex]

But that's not right... (I drew up a graph and did some calculations, it's supposed to be 0.36-something)

Then I thought I could make it equal to
[tex]e^{lim (3-n)ln\frac{n+2}{n+1}}[/tex]
so I could get rid of the (3-n) power, but then I would end up with infinity * 0 on top of the e.

A clue as to what I should be doing please? ^^ Thanks.
 

StatusX

Homework Helper
2,563
1
First I would change the variable from n to m-1. If m goes to infinity, n does as well, so you'll get the same limit. Then, do you know the formula for e? See if you can put this in a similar form.
 
639
2
use exponent rules to reduce the original expression. [tex] X^{a+b}= X^a X^b [/tex] and then try it.
 
57
0
Ok using what both of you said...

if n = m-1

[tex](1 + \frac{1}{m}) ^ {2-m}[/tex]

and using an exponent rule to get

[tex]\frac {(1 + \frac{1}{m}) ^ 2}{(1 + \frac{1}{m}) ^ m}[/tex]
[tex]= \frac {1}{e}[/tex]

... which seems to be the answer... thanks very much StatusX and Gale17 :)
 
57
0
oops sorry, i'm missing the lim as m->infinity signs...

well if you put those in, it makes sense :)
 

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