On one processor we have

$$T_1 = (1-f)W + fW = W$$

On $p$ processors we have

$$T_p = (1-f)W + \frac{fW}{p},$$

resulting in a speedup of

$$\frac{T_1}{T_p} = \frac{W}{(1-f)W+fW/p}$$

As $p$ goes to infinity, $fW/p$ goes to zero, and the maximum speedup is

$$\frac{1}{1-f},$$

meaning that if if $f = 0.99$ (all but $1\%$ parallelizable), the maximum speedup is $1/(1-.99)=100$!