Question

How do you find the limit of `ln(1+x)-lnx` as `x->oo`?

Answer 1

`0`

Explanation:

`ln(1+x)-lnx=ln((1+x)/x)=ln(1+1/x)`

as `x->oo` , `color(white)(88)1+1/x->1`

`ln(1)=0`

`:.`

`lim_(x->oo)ln(1+1/x)=0`