If we let `g(x)=ln(x+1)` and `h(x)=ln x,` we can find the limit of the function `f(x)=(ln(x+1))/ln x=(g(x))/(h(x))` as `x->oo` using L'Hopital's rule, and the limit of the corresponding sequence will of course be the same. The numerator and denominator both approach `oo` as `x` does, so `f` is indeed a...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

If we let `g(x)=ln(x+1)` and `h(x)=ln x,` we can find the limit of the function `f(x)=(ln(x+1))/ln x=(g(x))/(h(x))` as `x->oo` using L'Hopital's rule, and the limit of the corresponding sequence will of course be the same. The numerator and denominator both approach `oo` as `x` does, so `f` is indeed a candidate to try L'Hopital. Since

`g'(x)=1/(x+1)` and `f'(x)=1/x,` we get

`lim_(x->oo)f(x)=lim_(x->oo)(g(x))/(h(x))=lim_(x->oo)(g'(x))/(h'(x))=lim_(x->oo)(1/(x+1))/(1/x)=lim_(x->oo)x/(x+1)=1.`

**The limit is `1.` **

**Further Reading**