Stopping time of Feller process 
Let $X$ be a Feller process on $\mathbb{R}$ with generator $Gf=\frac{1}{2}f''-f'$ on $C_c^2$. Let $\tau_b$ be the first time that $X$ hits $b\in\mathbb{R}$. Show that for $x > 0$, $bP_x(\tau_b < \tau_0)\to 0$ as $b\to\infty$ and compute $E_x[\tau_0]$.

The way I have solved problems similar to the second part is to use Doobs stopping theorem on some appropriate martingale. Indeed we even have a variety of martingales at our disposal thanks to the fact that $M_t=f(X_t)-\int_0^t Gf(X_s)\,ds$ is a martingale for $f\in C_c^2$. But I don't see an appropriate $f$ that would help me out here, neither for the first nor the second part. I would appreciate any hints.
 A: Hints for Part 1: Set $$\tau :=\tau_{0,b} := \min\{\tau_0,\tau_b\}.$$


*

*Set $g(x) := \tfrac{1}{2} \exp(2x)$. Use Dynkin's formula to show that $$\mathbb{E}^x(g(X_{t \wedge \tau}))-x = \mathbb{E}^x \left( \int_0^{t \wedge \tau} (Gg)(X_s) \, ds \right).$$

*As $Gg(x)=0$ it follows from Step 1 that $$\mathbb{E}^x g(X_{t \wedge \tau}) = g(x).$$ Apply the dominated convergence theorem to conclude that $$\mathbb{E}^x g(X_{\tau})=g(x).$$

*Show that $$\mathbb{E}^x g(X_{\tau}) = \frac{1}{2} \left( e^{2b} \mathbb{P}^x(\tau_b<\tau_0)+ \mathbb{P}^x(\tau_0<\tau_b) \right)$$ and deduce from Step 2 that $$\limsup_{b \to \infty} e^{2b} \mathbb{P}^x(\tau_b<\tau_0)< \infty;$$ hence $$\lim_{b \to \infty} b \mathbb{P}^x(\tau_b<\tau_0)=0.$$
Hints for Part 2: Set $$\tau :=\tau_{0,b} := \min\{\tau_0,\tau_b\}.$$


*

*Set $f(x) := x$. Use Dynkin's formula to show that $$\mathbb{E}^x(f(X_{t \wedge \tau}))-x = \mathbb{E}^x \left( \int_0^{t \wedge \tau} (Gf)(X_s) \, ds \right).$$

*Since $Gf(x)=-1$ it follows that $$\mathbb{E}^x f(X_{t \wedge \tau})-x = - \mathbb{E}^x (t \wedge \tau).$$

*Use $|f(X_{t \wedge \tau})| \leq b$ and the monotone convergence theorem to prove that $\mathbb{E}^x(\tau)<\infty$.

*Conclude from Step 2 and 3 that $$b \mathbb{P}^x(\tau_b<\tau_0)-x = \mathbb{E}^x f(X_{\tau})-x = - \mathbb{E}^x(\tau). \tag{1}$$

*Use Part I and the monotone convergence theorem to let $b \to \infty$ in $(1)$. Conclude that $\mathbb{E}^x(\tau_0)=x$.


Remark: The process $X$ is actually a Brownian motion with drift, $$X_t = B_t-t; $$ this follows from the particular form of the generator.
