Why does expanding $\lim_{p \rightarrow 0}\frac{1-p-(1-p)^3}{1-(1-p)^3}$ give a different limit from just substituting? $$\lim_{p \rightarrow 0}\frac{1-p-(1-p)^3}{1-(1-p)^3}$$
$$= \frac{1-0-(1-0)^3}{1-(1-0)^3}$$
$$=\frac{0}{0}$$

$$\lim_{p \rightarrow 0}\frac{1-p-(1-p)^3}{1-(1-p)^3}$$
$$=\lim_{p \rightarrow 0}\frac{p^2-3p+2}{p^2-3p+3}$$
$$=\frac{0^2-3(0)+2}{0^2-3(0)+3}$$
$$=\frac{2}{3}$$
Edit 1: 
$0\le p \le1$

You conveniently omitted the crucial step
$$\lim_{p \rightarrow 0}\frac{1-p-(1-p)^3}{1-(1-p)^3}$$
$$=\color{red}{\lim_{p \rightarrow 0}\frac{p^3-3p^2+2p}{p^3-3p^2+3p}}$$
$$=\color{red}{\lim_{p \rightarrow 0}\frac pp} \cdot\lim_{p \rightarrow 0}\frac{p^2-3p+2}{p^2-3p+3}$$
$$=\frac{0^2-3(0)+2}{0^2-3(0)+3}$$
$$=\frac{2}{3}$$
The expression in red is clearly of the form $\dfrac 00$ too but will not be any more after you divide the numerator and the denominator by $p$. Functions that behave like that are said to have removable singularities.
