Exponential problem

IvanCarrie

New member
when going from here

$\lim_{n\rightarrow \infty} n \bigg(\bigg[1-\frac{1+\epsilon}{\frac{n}{\ln n}}\bigg]^{\frac{n}{\ln n}} \bigg)^{\frac{(n-1) \ln n}{n}}$

to here

$=\lim_{n\rightarrow \infty} n e^{-(1+\epsilon)\ln n}$

I can't see how the
$\frac{(n-1) \ln n}{n}$
becomes
$\ln n$

Subhotosh Khan

Super Moderator
Staff member
when going from here

$\lim_{n\rightarrow \infty} n \bigg(\bigg[1-\frac{1+\epsilon}{\frac{n}{\ln n}}\bigg]^{\frac{n}{\ln n}} \bigg)^{\frac{(n-1) \ln n}{n}}$

to here

$=\lim_{n\rightarrow \infty} n e^{-(1+\epsilon)\ln n}$

I can't see how the
$\frac{(n-1) \ln n}{n}$
becomes
$\ln n$
(n - 1)/n = 1 - 1/n

When the limit is calculated for n $$\displaystyle \to \ \infty$$, we get 0 for 1/n