Determining Properties of Stochastic Differential Equation as Drift Tends to $0$ Consider the stochastic differential equation
$$dS=\left(\mu S+\frac{\Lambda-S}{\omega}\right)dt+\sigma S dW_t.$$
such that $W_t$ is a Wiener process and $\mu,\sigma,\Lambda,\omega\in\mathbb{R}^+$. As $\omega\rightarrow 0$, does $S_t\rightarrow\Lambda$ for all $\mu$ (or is $S\sim\Lambda$)? If the expected value of $S_t$ was given by drift, $\frac{dS}{dt}=\frac{\Lambda-S_t}{\omega}+\mu$, then $\mathbb{E}[S_t]=c_0e^{-t/\omega}+\Lambda+\mu\omega$ and $\mathbb{E}[S_t]\rightarrow\Lambda$ as $\omega\rightarrow0$. Any help would be much appreciated.
Response to Kurt G.
Suppose
\begin{align}
S_t&=\Phi_t\Bigg(S_0-\frac{\Lambda}{\omega}\int_0^t\frac{1}{\Phi_s}\,ds\Bigg),\\\
\Phi_t&:=e^{\sigma W_t\,-\,\sigma^2t/2\,+\,\alpha t}.\
\end{align}
If $W_t=0$ for all $t$ (to make our calculation much simpler), then
\begin{align}
\int_0^t \frac{1}{\exp\left(\alpha s-\frac{\sigma^2 s}{2}\right)} ds=-\frac{2\left(e^{\frac{1}{2}t\left(\sigma^{2}-2\alpha\right)}-1\right)}{2\alpha-\sigma^{2}}.
\end{align}
Hence
\begin{equation}
\begin{split}
S&=e^{\alpha t-\frac{\sigma^2t}{2}}\left(S_0-\frac{\Lambda}{\omega}\left(-\frac{2\left(e^{\frac{1}{2}t\left(\sigma^2-2\alpha\right)}-1\right)}{2\alpha-\sigma^2}\right)\right),\\
&=e^{\alpha t-\frac{t\sigma^2}{2}}\left(S_0+\frac{2\Lambda\left(e^{\frac{t\left(\sigma^2-2\alpha\right)}{2}}-1\right)}{\omega\left(2\alpha-\sigma^2\right)}\right),\\
&=e^{\alpha t-\frac{t\sigma^2}{2}}S_0+\frac{2\Lambda\left(1- e^{\frac{2\alpha t-t\sigma^2}{2}}\right)}{\omega(2\alpha-o^2)}.
\end{split}
\end{equation}
So, as $\omega\rightarrow 0$, $S\rightarrow -\Lambda$. This seems counterintuitive since $S$ should be increasing when $S<\Lambda$ and decreasing when $S>\Lambda$. So, where is the mistake?
 A: For $\Lambda=0$ the answer is fairly simple because then the SDE is the familiar Black-Scholes SDE
$$
\frac{dS_t}{S_t}=\alpha\,dt+\sigma\,dW_t\,,\quad\quad\alpha=\mu-\frac{1}{\omega}\,,
$$
which has the solution
$$
S_t=S_0\,e^{\sigma W_t-\,\sigma^2t/2\,+\,\alpha\, t}\,.
$$
Clearly $E[S_t]=S_0\,e^{\alpha t}$ which converges to $\Lambda=0$ when
when $\omega\to 0$ because $\alpha\to-\infty\,.$ To figure out what happens pathwise we can look at
$$
E[S_t^2]=S_0^2\,e^{\sigma^2t\,+\,2\,\alpha\,t}\,.
$$
Obviously, for every fixed $t$ this squared $L_2$-norm converges to zero when $\omega\to 0\,.$
Therefore, $S_t$ converges almost surely to $\Lambda=0\,.$
In general, maybe the following helps: The explicit solution to your SDE is
\begin{align}
S_t&=\Phi_t\Bigg(S_0+\frac{\Lambda}{\omega}\int_0^t\frac{1}{\Phi_s}\,ds\Bigg)\,,\\
\Phi_t&:=e^{\sigma W_t\,-\,\sigma^2t/2\,+\,\alpha t}\,.
\end{align}
(see Wikipedia). This is easily verified. Because $\Phi_t$ is identical to $S_t$ when $\Lambda$ is zero we know from above that $\Phi_t$ converges almost surely to zero when $\omega\to 0\,.$
To figure out what $S_t$ is doing when $\omega\to 0$ observe that
$$\tag{1}
\Phi_t\int_0^t\frac{1}{\Phi_s}\,ds=e^{\sigma W_t\,-\,\sigma^2t/2}\int_0^te^{-\sigma W_s\,+\,\sigma^2s/2\,+\,\alpha(t-s)}\,ds\,.
$$
Since $e^{\alpha(t-s)}$ converges to zero when $\omega\to 0$ I think it follows from the dominated convergence theorem that the expresson (1) converges to zero. It remains to show that
$$
\frac{\Phi_t}{\omega}\int_0^t\frac{1}{\Phi_s}\to 1\,.
$$
