Derivation of the Shannon Limit
Your reasoning is almost correct.

In the step where you take the limit of the \${E_b}/{N_0}\$ equation, you're making a mistake when applying L'Hôpital's rule. It should go as follows:

$$
\lim_{C/W \to 0^+} \frac{E_b}{N_0} = \lim_{C/W \to 0^+} \frac{2^{C/W} - 1}{C/W} = \lim_{C/W \to 0^+} \ln {(2)} 2^{C/W} = \ln{(2)}
$$

Your problem arises when you take the derivative of \$2^{C/W} - 1\$, just remember that

$$
\frac{d(a^{x})}{dx} = \ln{(a)} a^{x}
$$

Hope it helps.