Take the 2-minute tour ×
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|improve this question
add comment

3 Answers

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|improve this answer
    
Thanks very much for help. Really helpful.thanks –  Human love Oct 8 '11 at 5:18
add comment

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|improve this answer
add comment

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|improve this answer
add comment

Your Answer

 
discard

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.