Here's a way to look at the second one if you like.
Use the series for $\displaystyle \frac{ln(1+x)}{x}=1-\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+\frac{x^{4}}{5}-\frac{x^{5}}{6}+......$
$\displaystyle \lim_{x\to 0}\left[1-\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+\frac{x^{4}}{5}-\frac{x^{5}}{6}+......\right]$
See now where the 1 comes in?. As all the x terms drop due to the 0, all that's left is 1.