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.
please help!
 A: 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.
A: 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 
$$\lim_{x\rightarrow \infty} \left(\frac{\ln(x)}{x}\right)^\frac{1}{x} =e^0 =1$$
A: 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 $L$.
