Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Please help me in evaluating following limit:

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

share|cite|improve this question

3 Answers 3

up vote 4 down vote accepted

Let's denote

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

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

share|cite|improve this answer
Thanks very much for help. Really helpful.thanks – Human Love Oct 8 '11 at 5:18

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}$.

share|cite|improve this answer

Let the value of the limit be y.

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.