- #1
shan
- 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.
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.