How would I use the definition of derivative to prove $$\lim_{x\to 0} \frac{\ln(1+x)}{x} = 1$$
I got to $$\frac{\frac{\ln(1+x+h)}{(x+h)} - \frac{\ln(1+x)}{x}}{h}$$ but have no idea where to go from here.
On another site I found someones answer where they stated the following: $$ \lim_{x\to 0} \frac{\ln(1+x) - \ln(1+0)}{x-0} = [\ln(1+x)]'\rvert_x = 0 $$ but I am unsure why the $x$ in the $x-0$ is removed. Can someone please explain?