Question

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

Answer 1

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`