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Answer 1 · 2017-02-03T00:41:35

$lim_(x->oo)ln(ln(x))/sqrt(x) = 0$

$lim_(x->oo)ln(ln(x))/sqrt(x)$ produces an $oo/oo$ indeterminate form on direct substitution, and so we may apply L'Hopital's rule.

$lim_(x->oo)ln(ln(x))/sqrt(x) = lim_(x->oo)(d/dxln(ln(x)))/(d/dxsqrt(x))$

$=lim_(x->oo)(1/(xln(x)))/(1/(2sqrt(x))$

$=lim_(x->oo)2/(sqrt(x)ln(x))$

$=2/oo$

$=0$

Answer 2 · 2017-02-03T00:54:43

$0$

$e^x$ is a monotonically increasing function so

$lim_(x->oo)log(log(x))/sqrt(x)=lim_(x->oo)e^(log(logx))/e^sqrt(x)=lim_(x->oo)log(x)/e^sqrt(x)$

Here $log(x) < x$ and $e^sqrt(x)$ grows much more quickly than any polynomial function so

$lim_(x->oo)log(log(x))/sqrt(x)=lim_(x->oo)log(x)/e^sqrt(x)=0$