Lim as n->infinity help

1. Jun 17, 2005

shan

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

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

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$$(1 + \frac{1}{n+1})$$ = 1

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

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.

2. Jun 18, 2005

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.

3. Jun 18, 2005

Gale

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

4. Jun 18, 2005

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 :)

5. Jun 18, 2005

shan

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

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