Best Answer
1 method
intutive
as ln (x) grows much smaller than x and even ln(ln x) even slower so
ln (ln x)/x = 0 as x>infinite
2) method
let ln x = y as x >inf y > inf
we have ln y/e^y
of the form inf/inf so using L hosplital; rlue
1/y/(e^y) as d/dy(ln y) = 1/y and d/dy(e^y) = e^y]
= 1/(ye^y)
= 0
intutive
as ln (x) grows much smaller than x and even ln(ln x) even slower so
ln (ln x)/x = 0 as x>infinite
2) method
let ln x = y as x >inf y > inf
we have ln y/e^y
of the form inf/inf so using L hosplital; rlue
1/y/(e^y) as d/dy(ln y) = 1/y and d/dy(e^y) = e^y]
= 1/(ye^y)
= 0
Other Answers (1)

As x approaches infinty the numerator ln(ln(x)) gets very large.
The denominator x also get very large.
This produces the indeterminate form infinity/infinite
Using the L'hospital Rule take the derivative of the numerator divided by the derivative of the denominator
Then take the limit as x approaches infinity of the result.
Result: 1/(x*ln(x))
Let f(x) = 1/(x*ln(x))
The limit of f(x) as x approaches infinity is zero since
the numerator (1) is smaller than the denominator (x*ln(x)) as
x gets larger.