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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#