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

1 Answer

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

Explanation:

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