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Please help me in evaluating following limit:

$$\lim_{n \to \infty}\left(\frac1{n}\right)^\frac1{\ln n}$$

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up vote 4 down vote accepted

Let's denote

$A=\left(\frac1{n}\right)^\frac1{\ln n} \Rightarrow \ln A=\ln \left(\frac1{n}\right)^\frac1{\ln n} \Rightarrow \ln A=\frac{1}{\ln n}\ln \frac{1}{n} \Rightarrow \ln A=\frac{-1}{\ln n}\ln n\Rightarrow \ln A=-1 \Rightarrow$

$\Rightarrow A=e^{-1} \Rightarrow \lim_{n \to \infty}\left(\frac1{n}\right)^\frac1{\ln n}=\lim_{n \to \infty} e^{-1}=e^{-1}$

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Thanks very much for help. Really helpful.thanks – Human Love Oct 8 '11 at 5:18

Let the value of the limit be y.

$$\log y = \lim_{n \to \infty}(1/\log n)(-\log n) = -1$$ So $y = 1/e$

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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}$.

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