Perhaps someone could explain the last step here for me please...

Find the limit of $\displaystyle \mathop {\lim}\limits_{x \to 1} \frac{In(x)}{x-1}$. Do not use L'Hopital's rule.

Solution;

Since In(1) = 0, we have that,

$\displaystyle \mathop {\lim}\limits_{x \to 1} \frac{In(x)}{x-1}$ = $\displaystyle \mathop {\lim}\limits_{x \to 1} \frac{In(x) - In(1)}{x-1}$ = $\displaystyle \frac{d}{dx} In(x) = 1$ for x =1. (Not quite sure how to do the latex for the way its written in my notes. To the right of the d/dx is a vertical line with x=1 at the bottom.)

And... Im not quite sure how to do the last step there i.e the $\displaystyle \mathop {\lim}\limits_{x \to 1} \frac{In(x) - In(1)}{x-1}$ = $\displaystyle \frac{d}{dx} In(x) = 1$ bit...

Any help on this?