Lim as x --> infinity of ln(lnx) / x?

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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
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Other Answers (1)

  • ≈ nohglf answered 6 years ago
    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.
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