You need to evaluate the limit of function `ln(1+3x)/x`  if `x -gt0` , such that:

`lim_(x-gt0) ln(1+3x)/x`

Substituting 0 for x yields:`lim_(x-gt0) ln(1+3x)/x = ln(1+0)/0 = (ln1)/0 =gt lim_(x-gt0) ln(1+3x)/x = 0/0`

This kind of indetermination lets you to use l'Hospital's theorem such that:

`lim_(x-gt0) ln(1+3x)/x = lim_(x-gt0) ((ln(1+3x))')/(x')`

`` `lim_(x-gt0) ln(1+3x)/x = lim_(x-gt0) ((1+3x)'/(1+3x))/1 `

`lim_(x-gt0) ln(1+3x)/x = lim_(x-gt0) 3/(1+3x)`

`lim_(x-gt0) 3/(1+3x) = 3/(1+0) = 3`

Hence, evaluating the limit to the function `ln(1+3x)/x , x-gt 0` , yields:`lim_(x-gt0) ln(1+3x)/x = 3` .