To find the limit of [tex]\(\left( \frac{\ln x}{x} \right)^{1 / \ln x}\)[/tex] as [tex]\(x \to \infty\)[/tex], we need to carefully analyze the behavior of the expression as [tex]\(x\)[/tex] grows larger.
[tex]\[ \lim _{x \rightarrow \infty}\left(\frac{\ln x}{x}\right)^{1 / \ln x} \][/tex]
First, let's denote the given expression as [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = \left( \frac{\ln x}{x} \right)^{1 / \ln x} \][/tex]
Taking the natural logarithm of [tex]\( f(x) \)[/tex] will help simplify the analysis. Let:
[tex]\[ L = \ln(f(x)) = \ln \left( \left( \frac{\ln x}{x} \right)^{1 / \ln x} \right) \][/tex]
Using the logarithm power rule [tex]\( \ln(a^b) = b\ln(a) \)[/tex]:
[tex]\[ L = \frac{1}{\ln x} \ln \left( \frac{\ln x}{x} \right) \][/tex]
Next, we can split the logarithm inside:
[tex]\[ L = \frac{1}{\ln x} \left( \ln (\ln x) - \ln x \right) \][/tex]
Separate the terms in the fraction:
[tex]\[ L = \frac{\ln (\ln x)}{\ln x} - \frac{\ln x}{\ln x} \][/tex]
[tex]\[ L = \frac{\ln (\ln x)}{\ln x} - 1 \][/tex]
Now, we are interested in finding the limit of [tex]\( L \)[/tex] as [tex]\( x \to \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} L = \lim_{x \to \infty} \left( \frac{\ln (\ln x)}{\ln x} - 1 \right) \][/tex]
Consider each term separately. First, observe the fraction [tex]\(\frac{\ln (\ln x)}{\ln x}\)[/tex]:
As [tex]\( x \to \infty \)[/tex], both [tex]\( \ln x \)[/tex] and [tex]\( \ln(\ln x) \)[/tex] increase, but [tex]\( \ln x \)[/tex] increases much faster than [tex]\( \ln(\ln x) \)[/tex]. Hence, we can conclude:
[tex]\[ \lim_{x \to \infty} \frac{\ln (\ln x)}{\ln x} = 0 \][/tex]
Thus, the limit of [tex]\( L \)[/tex] becomes:
[tex]\[ \lim_{x \to \infty} L = 0 - 1 = -1 \][/tex]
We have determined that:
[tex]\[ \ln(f(x)) \to -1 \][/tex]
Finally, to find the limit of [tex]\( f(x) \)[/tex], we exponentiate both sides:
[tex]\[ f(x) \to e^{-1} \][/tex]
Therefore, the limit is:
[tex]\[ \lim_{x \to \infty} \left( \frac{\ln x}{x} \right)^{1 / \ln x} = e^{-1} \][/tex]
Thus, the answer is:
[tex]\[ e^{-1} \][/tex]
