What is the limit of #(x/(x+1))^x# as x approaches infinity?

1 Answer
Konstantinos Michailidis
Feb 22, 2016

It is #1/e#

Explanation:

We can rewrite this as

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

Hence we need to find the limit of

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

We notice that the denominator and the numerator goes to zero as x goes to infinity hence we can apply L'Hospital rule since we have a #0/0# indeterminate form.

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

Hence the initial limit is #e^(-1)=1/e#

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