How I resolve this limit?

#lim_(n->+oo)(1/logn)^((log(logn))/n)#

THANKS!

1 Answer
Dean R.
Jun 22, 2018

#1#

Explanation:

#lim_(n->+oo)(1/ln n)^((ln ln n)/n)#

#lim_(n->+oo)e^ { ln( (1/ln n)^((ln ln n)/n) )}#

#= exp( lim_(n->+oo) ((ln ln n)/n) ln( 1/ln n) )#

#= exp( lim_(n->+oo) -(ln ln n)^2/n)#

#= exp( 0 )#

# = 1#

I wrote #exp(x)# instead of #e^x# because the latter doesn't look very good with the entire limit in the exponent. #exp(x)=e^x#

In the limit of a very slowly increasing #(ln ln n)^2# in the numerator versus a must faster increasing #n# in the denominator, the denominator wins and the limit is zero.

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