How do you find the limit of #(x/(x+1))^x# as x approaches #oo#?

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Answer 1 · 2016-06-12T22:49:57

The limit is #l=e^-1=1/e#

we can write this as

#e^[x*ln(x/(1+x))]#

Hence actually we need to find the limit

#lim_(x->oo) [x*ln(x/(1+x))]=lim_(x->oo) [ln(x/(1+x))/(1/x)]#

Because

#ln(x/(1+x))/[1/x]->0/0# apply L'Hopital law hence

#([ln(x/(1+x))]')/([1/x]')=[1/(x^2+x)]/(-1/x^2)= -x^2/(x^2+x)=-x/(1+x)->-1# as #x->oo#

Hence the limit is #e^-1=1/e#