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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.