I think you nearly got there but made a few mistakes in the application of l'Hopital's rule.

First Limit

In the first case, you got

\begin{eqnarray} \lim_{S_0 \rightarrow \infty} \Omega & = & \lim_{S_0 \rightarrow \infty} \frac{\Gamma_{\text{call}} S_0 + \Delta_{\text{call}}}{\Delta_{\text{call}}}\\ & = & \lim_{S_0 \rightarrow \infty} \frac{\Gamma_{\text{call}} S_0 + 1}{1} \end{eqnarray}

and you seem to conclude that $\lim_{S_0 \rightarrow \infty} \Gamma_{\text{call}} S_0 = \infty$ and thus $\lim_{S_0 \rightarrow \infty} \Omega = \infty$. This is not true however. Remember that

\begin{equation} \Gamma_{\text{call}} = \frac{\mathcal{N}' \left( d_+ \right)}{S_0 \sigma \sqrt{T}} \end{equation}

and thus

\begin{equation} \lim_{S_0 \rightarrow \infty} \Gamma_{\text{call}} S_0 = \lim_{S_0 \rightarrow \infty} \frac{\mathcal{N}' \left( d_+ \right)}{\sigma \sqrt{T}} = 0. \end{equation}

Consequently $\lim_{S_0 \rightarrow \infty} \Omega_{\text{call}} = 1$ as postulated by the sample solution.

Second Limit

I agree with you up to the point where you have

\begin{equation} \lim_{S_0 \rightarrow 0} \Omega_{\text{call}} = \frac{\Gamma_{\text{call}} S_0 + \Delta_{\text{call}}}{\Delta_{\text{call}}} \end{equation}

which results in a $0 / 0$ situation. However, when you apply l'Hopital's rule again now, you don't apply the chain rule correctly and forgot to differentiate gamma. I get

\begin{equation} \ldots = \lim_{S_0 \rightarrow 0} \frac{\mathcal{S}_{\text{call}} S_0 + 2 \Gamma_{\text{call}}}{\Gamma_{\text{call}}} \end{equation}

where I use $\mathcal{S}_{\text{call}}$ to denote the third derivative w.r.t. the spot (the speed). It is given by

\begin{equation} \mathcal{S}_{\text{call}} = -\frac{\Gamma_{\text{call}}}{S} \left( \frac{d_+}{\sigma \sqrt{T}} + 1 \right). \end{equation}

We thus get

\begin{equation} \ldots = \lim_{S_0 \rightarrow 0} \left\{ -\left( \frac{d_+}{\sigma \sqrt{T}} + 1 \right) + 2 \right\}. \end{equation}

Now, since $\lim_{S_0 \rightarrow 0} d_+ = -\infty$, this yields $\lim_{S_0 \rightarrow 0} \Omega_{\text{call}} = \infty$ as postulated by the sample solution.

I think you nearly got there but made a few mistakes in the application of l'Hopital's rule.

First Limit

In the first case, you got

\begin{eqnarray} \lim_{S_0 \rightarrow \infty} \Omega & = & \lim_{S_0 \rightarrow \infty} \frac{\Gamma_{\text{call}} S_0 + \Delta_{\text{call}}}{\Delta_{\text{call}}}\\ & = & \lim_{S_0 \rightarrow \infty} \frac{\Gamma_{\text{call}} S_0 + 1}{1} \end{eqnarray}

and you seem to conclude that $\lim_{S_0 \rightarrow \infty} \Gamma_{\text{call}} S_0 = \infty$ and thus $\lim_{S_0 \rightarrow \infty} \Omega = \infty$. This is not true however. Remember that

\begin{equation} \Gamma_{\text{call}} = \frac{\mathcal{N}' \left( d_+ \right)}{S_0 \sigma \sqrt{T}} \end{equation}

and thus

\begin{equation} \lim_{S_0 \rightarrow \infty} \Gamma_{\text{call}} S_0 = \lim_{S_0 \rightarrow \infty} \frac{\mathcal{N}' \left( d_+ \right)}{\sigma \sqrt{T}} = 0. \end{equation}

Consequently $\lim_{S_0 \rightarrow \infty} \Omega_{\text{call}} = 1$ as postulated by the sample solution.

Second Limit

I agree with you up to the point where you have

\begin{equation} \lim_{S_0 \rightarrow 0} \Omega_{\text{call}} = \frac{\Gamma_{\text{call}} S_0 + \Delta_{\text{call}}}{\Delta_{\text{call}}} \end{equation}

which results in a $0 / 0$ situation. However, when you apply l'Hopital's rule again now, you don't apply the chain rule correctly and forgot to differentiate gamma. I get

\begin{equation} \ldots = \lim_{S_0 \rightarrow 0} \frac{\mathcal{S}_{\text{call}} S_0 + 2 \Gamma_{\text{call}}}{\Gamma_{\text{call}}} \end{equation}

where I use $\mathcal{S}_{\text{call}}$ to denote the third derivative w.r.t. the spot (the speed). It is given by

\begin{equation} \mathcal{S}_{\text{call}} = -\frac{\Gamma_{\text{call}}}{S_0} \left( \frac{d_+}{\sigma \sqrt{T}} + 1 \right). \end{equation}

We thus get

\begin{equation} \ldots = \lim_{S_0 \rightarrow 0} \left\{ -\left( \frac{d_+}{\sigma \sqrt{T}} + 1 \right) + 2 \right\}. \end{equation}

Now, since $\lim_{S_0 \rightarrow 0} d_+ = -\infty$, this yields $\lim_{S_0 \rightarrow 0} \Omega_{\text{call}} = \infty$ as postulated by the sample solution.