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Since Shroomery doesn't have a math forum, I think this would be the best forum for me to ask for homework help. (I'm not asking for you to give me the work straight up, just any tips, so please don't flame me)
The assigned problem is Prove using L'Hopital's Rule that lim (n->infinity) (1 + 1/n)^n = e.
I first tried combining the term to get:
lim [(n+1)/n^2]^n = lim (n+1)^n/n^(2n)
but after taking the derivative (as per L'Hopital's Rule), I couldn't figure out anything, and enough of the terms didn't cancel.
So I tried to get the equality another way (not using L'Hopital's Rule) by setting
ln y = ln lim (n->infinity) (1 + 1/n)^n
ln y = n * lim ln (1 + 1/n) --- ln 1/n = 0, so it's just ln 1 = 0
ln y = n * 0 = 0, but this cannot exist.
Can somebody be kind enough to tell me where I'm going wrong or if there's a better method for me to use (preferably using L'Hopital's Rule)? Any help is definitely appreciated!
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Found it. I'll use x instead of n. inf. = "infinity"
[lim x to inf.] (1 + 1/x)^x
We have to change this to the form 0 / 0 or inf. / inf.
let y = (1 + 1/x)^x
thus [lim x to inf.] ln y = [lim x to inf.] x ln (1 + 1/x)
divide by (1/x) / (1/x)
so [lim x to inf.] ln y = [lim x to inf.] < ln (1 + 1/x) > / ( 1 / x)
which is of the form 0 / 0
apply L'Hopital's theorem
[lim x to inf.] ln y = [lim x to inf.] < 1 / (1 + 1/x) * ( -1 / x^2 ) > / ( -1 / x^2)
the ( -1 / x^2) terms cancel out, leaving
[lim x to inf.] ln y = [lim x to inf.] 1 / (1 + 1/x)
[lim x to inf.] e^(ln y) = [lim x to inf.] e ^< 1 / (1 + 1/x) >
[lim x to inf.] y = e^1
since y = (1 + 1/x)^x
[lim x to inf.] (1 + 1/x)^x = e
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