Question

What is the limit of `ln(x+1)/x` as x approaches `oo`?

Answer 1

Use L'Hôpital's rule. Answer is:

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

Explanation:

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

This limit can't be defined as it is in form of `oo/oo` Therefore you can find the derivative of the nominator and denumerator:

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

`=lim_(x->oo)(1/(x+1)*(x+1)')/1=lim_(x->oo)1/(x+1)*1=`

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

As you can see through the chart it indeed tends to approach `y=0`

graph{ln(x+1)/x [-12.66, 12.65, -6.33, 6.33]}