Question #5e925

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Answer 1 · 2017-11-17T14:39:20

The answer is #=1/e#

We need

#ln1=0#

Let rewrite and simplify

#a_n=(1/n)^(1/lnn)#

#ln(a_n)=1/lnnln(1/n)=1/lnn*(ln1-lnn)#

#=1/lnn*(0-lnn)=1/lnn(-lnn)=-1#

Therefore,

#lna_n=-1#

#a_n=e^-1=1/e#

So,

#lim_(n->oo)(1/n)^(1/lnn)=1/e#

graph{(1/x)^(1/lnx) [-2.79, 17.21, -4.92, 5.08]}