By Shannon-Nyquist theorem, a bandlimited signal can be completely reconstructed with $latex 2W$ samples per second. Think of the opposite way, this means that each second we have $latex 2W$ independent components that we can vary for each second. The total noise power per second is $latex 2 W (N_0/2) = N_0 W$ and as it is spread over $latex 2W$ channel, the noise per component is $latex N_0/2$. Similarly, if the power of the signal is $latex P$, each component will have $latex P/(2W)$. And each of the components have capacity $latex \frac{1}{2}\log_2 (1+\frac{P/(2W)}{N_0/2})=\frac{1}{2}\log_2 (1+\frac{P}{N_0 W})$. Now, since we have $latex 2W$ of such independent components (hence independent channels), the total capacity is $latex W\log_2(1+\frac{P}{N_0 W})$.
Band-unlimited capacity
What will be the capacity if we don’t have infinite bandwidth? By L’ Hopital’s rule,
$latex \lim_{W\rightarrow \infty}W\log_2(1+\frac{P}{N_0 W}) \mbox{bits}=\lim_{W\rightarrow \infty}\frac{\ln(1+\frac{P}{N_0 W})}{1/W} \mbox{nats}=\lim_{W\rightarrow \infty}\frac{\frac{1}{1+P/(N_0 W)}}{-1/W^2}\frac{P}{N_0}(-1/W^2) \mbox{nats}=\frac{P}{N_0} \mbox{nats}=\frac{P}{N_0} \log_2 e \; \mbox{bits}$.
Interestingly, even with infinite bandwidth, the capacity increases only linearly with the input power.