Exponential problem

IvanCarrie · Jan 2, 2019

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 · Jan 2, 2019

IvanCarrie said:
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

Exponential problem

IvanCarrie

New member

Subhotosh Khan

Super Moderator