Let $(B_t)$ a Brownian motion. Let $f:\mathbb R\to\mathbb R$ such that $f\in\mathcal C^2(\mathbb R)$ and such that $f''$ is bounded. Show that $$\lim_{h\to 0^+}\frac{\mathbb E[f(B_{t+h})\mid B_t=x]-f(x)}{h}=\frac{f''(x)}{2}.$$
In the proof, it's written that: by the Markov property of a Brownian motion, if $B_t^x$ is a brownien motion sucht that $B_0=x$, then
$$\mathbb E[f(B_{t+h})\mid B_t=x]\underset{(1)}{=}\mathbb E[f(B_h)\mid B_0=x]\underset{(2)}{=}\mathbb E[f(B_h^x)]\underset{(3)}{=}\mathbb E[f(x+B_h)].$$
I think that that fact that $f$ is continuous, $f(B_t)$ is also a brownian motion, no ? and by translation of the time, we have $(1)$. But I don't understand neither $(2)$ nor $(3)$.