Re: differentiation question (n^n and (n-1)^n)


Jules wrote:
ChineseBoy--love maths wrote:
Jules,
why lim n * (ln(n - 1) - ln(n)) = -1 ?

Write ln(n - 1) - ln(n) = ln( 1 - 1 / n), which goes to 0 as n -->
infinity.

So, lim n * (ln(n - 1) - ln(n)) = lim n * ln( 1 - 1 / n) = lim ln( 1 -
1 / n) / (1 / n).
At this point, the numerator and denominator both go to zero, so we use
l'Hospital's rule.

lim ln( 1 - 1 / n) / (1 / n) = lim [(1/(1 - 1/n)) * 1 / n ^ 2 ] / (-1 /
n^2) = lim -n / (n-1) = lim -1 - 1 / (n - 1) = -1.

Thanks for your reply.
I haven't studied it yet.

.

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