# Limit $\lim_{x\to\infty}\left(\frac{\ln x}x\right)^{1/x}$

I have to find the limit of the next thing:

$$\lim_{x\to\infty}\left(\frac{\ln x}x\right)^{1/x}$$

I think about: $y = \ln(x)$ and then $x = e^y$, but it will be so long.

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(ln(x))^(1/x) tends to 1 as x tends to infinity. Same with x^(1/x). So the limit is 1/1 = 1. – Adam Rubinson Feb 17 '13 at 12:43
This can be found step by step, using the Maple code $$with(Student[Calculus1]):LimitTutor((ln(x)/x)^{1/x},x=infinity);$$ – user64494 Sep 1 '13 at 17:32

Consider the limit of the logarithm first:

$$\lim_{x\to\infty} \ln\left(\dfrac{\ln x}{x}\right)^{1/x} = \lim_{x\to\infty} \frac{\ln\left(\dfrac{\ln x}{x}\right)}{x} = \lim_{x\to\infty} \frac{\ln \ln x - \ln x}{x}$$

Now apply L'Hôpital's rule:

$$\lim_{x\to\infty} \ln\left(\dfrac{\ln x}{x}\right)^{1/x} = \lim_{x\to\infty} \left(\dfrac{1}{x \ln x} - \frac{1}{x} \right) = 0$$

Thus, the original limit is $e^0 = 1$.

As @DominicMichaelis points out in the comments, taking the limit of the logarithm is justified because the logarithm is continuous and injective.

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As this is normally very elementary stuff, you should say why you can consider the limit of the logarithm. Here you need that the logarithm is injective and continuous. – Dominic Michaelis Feb 17 '13 at 13:14
@DominicMichaelis Indeed. Thanks for your note. – Ayman Hourieh Feb 17 '13 at 13:20

Let

$$L=\lim_{x\to\infty}\left(\frac{\ln x}x\right)^{1/x}\;.$$

Then by continuity of the log you have

$$\ln L=\lim_{x\to\infty}\ln\left(\frac{\ln x}x\right)^{1/x}=\lim_{x\to\infty}\frac1x(\ln\ln x-\ln x)=\lim_{x\to\infty}\frac{\ln\ln x}x-\lim_{x\to\infty}\frac{\ln x}x\;,$$

provided that the limits in question exist. Now apply l’Hospital’s rule to find $\ln L$, and exponentiate to find $L$.

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Just write it from $a^b$ to $e^{ln(a) \cdot b}$ and use the continuous of the e function and L'hospital. $$\left(\frac{\ln(x)}{x}\right)^\frac{1}{x}=e^{\frac{1}{x} \cdot (\ln(\ln(x))-\ln(x))}$$ Using L'hospital we have $$\lim_{x \rightarrow \infty} \frac{1}{x} \cdot (\ln(\ln(x))-\ln(x))= \lim_{x\rightarrow \infty} \frac{1}{x\cdot \log{x}} -\frac{1}{x}=0$$ And so $$\lim_{x\rightarrow \infty} \left(\frac{\ln(x)}{x}\right)^\frac{1}{x} =e^0 =1$$

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