Maple says that this limit is zero but I can't prove it. Any help or tips would be appreciated.
$\displaystyle\lim_{n\rightarrow\infty}n\left(1-\frac{1}{\ln(n)}\right)^n$
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Sign up to join this communityMaple says that this limit is zero but I can't prove it. Any help or tips would be appreciated.
$\displaystyle\lim_{n\rightarrow\infty}n\left(1-\frac{1}{\ln(n)}\right)^n$
We have
$$n\left(1-\frac{1}{\ln(n)}\right)^n=n\exp\left(n\ln\left(1-\frac{1}{\ln(n)}\right)\right)\sim_\infty n\exp\left(-\frac{n}{\ln n}\right)\xrightarrow{n\to\infty}0$$ and the last limit can be proved using the L'Hôpital's rule.
$$\lim_{n\to\infty}n\bigg(1-\frac{1}{\ln(n)}\bigg)^n=\lim_{n\to\infty}n\bigg[\bigg(1-\frac{1}{\ln(n)}\bigg)^{\ln(n)}\bigg]^{\frac{n}{\ln(n)}}=\lim_{n\to\infty}\frac{n}{e^{\frac{n}{\ln(n)}}}=0$$ Where I used the fact that $$\lim_{a_n\to\infty}\left(1-\frac{1}{a_n}\right)^{a_n}=e^{-1}$$ and for $\ n\to\infty$ $$\ n>>\ln(n)$$