What is the limit as n approaches infinity for (1 + \frac{1}{n+1}) ^ {3-n}?

[shan]

It seems to an area I'm really fuzzy on. Anyway, the question I've having problems with is

lim as n->infinity<br /> (1 + \frac{1}{n+1}) ^ {3-n}<br />

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<br /> (1 + \frac{1}{n+1})<br /> = 1

so lim as n->infinity<br /> (1 + \frac{1}{n+1}) ^ {3-n} = 1 ^ {3-n} = 1<br />

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
e^{lim (3-n)ln\frac{n+2}{n+1}}
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]

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.

 

[Gale]

use exponent rules to reduce the original expression. X^{a+b}= X^a X^b and then try it.

 

[shan]

Ok using what both of you said...

if n = m-1

(1 + \frac{1}{m}) ^ {2-m}

and using an exponent rule to get

\frac {(1 + \frac{1}{m}) ^ 2}{(1 + \frac{1}{m}) ^ m}
= \frac {1}{e}

... which seems to be the answer... thanks very much StatusX and Gale17 :)

 

[shan]

oops sorry, I'm missing the lim as m->infinity signs...

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