Please help me in evaluating following limit:
$\lim_{n \to \infty}\left(\frac1{n}\right)^\frac1{\ln n}$
Please help me in evaluating following limit:
$\lim_{n \to \infty}\left(\frac1{n}\right)^\frac1{\ln n}$
Let's denote
$A=\left(\frac1{n}\right)^\frac1{\ln n} \Rightarrow \ln A=\ln \left(\frac1{n}\right)^\frac1{\ln n} \Rightarrow \ln A=\frac{1}{\ln n}\ln \frac{1}{n} \Rightarrow \ln A=\frac{-1}{\ln n}\ln n\Rightarrow \ln A=-1 \Rightarrow$
$\Rightarrow A=e^{-1} \Rightarrow \lim_{n \to \infty}\left(\frac1{n}\right)^\frac1{\ln n}=\lim_{n \to \infty} e^{-1}=e^{-1}$
Let the value of the limit be y.
$\log y = \lim_{n \to \infty}(1/\log n)(-\log n) = -1$ So $y = 1/e$
Similar, but without logging:
$\lim_{n \to \infty}\left(\frac1{n}\right)^\frac1{\ln n} = \lim_{n \to \infty}\left(e^{\ln(1/n)}\right)^\frac1{\ln n} = \lim_{n \to \infty}\left(e^{-\ln(n)}\right)^\frac1{\ln n} = \lim_{n \to \infty}\left(e^{-\ln(n)\frac1{\ln n}}\right) $ $ = e^{-1}$.