Posted by Bash on Sunday, February 5, 2012 at 1:15am.
Rewrite as
Lim ln(1 + 1/n)/(1/n)
(n -> infinity)
Then use the power series for ln(1+x) for small x:
ln(1 + 1/n) = 1/n -(1/2)(1/n)^2 +(1/3)(1/n)^3...
Divide that by 1/n and you get 1 plus a series of higher order terms that become zero in the limit as n-> infinity.
The limit is 1.
Check: Try n = 100
100*(ln101 -ln100) = 0.9950
n = 1000:
1000*(ln1001 - ln1000) = 0.9995
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