Bandwidth-Data rate relationship
Reword Case II like this:


  Now suppose \$f=2\ \mathrm{MHz}\$. Then, using the same line of reasoning as before, the bandwidth of the signal is &lt;...equation...> = 8 MHz. But in this case \$T = 1/f = 0.5\ \mu\mathrm{s}\$. As a result, one bit occurs every 0.25 us for a rate of 4 Mbps. Thus other things being equal, by doubling the bandwidth we double the potential bit rate. 


In the original version they kind of put the cart before the horse by "supposing" the higher bandwidth, instead of starting with the frequency and showing how that implies a higher bandwidth.