Taylor's formula - infinitesimal generator

Question

As it is written here, using the Taylor's expansion, one can write

$$ \mathbb{E}^xf(X_t) \approx f(x)+t Af(x) $$ from $$ Af(x) := \lim_{t \downarrow 0} \frac{\mathbb{E}^x(f(X_t))-f(x)}{t} $$ where $A$ is the infinitesimal generator and $\mathbb{E}^x$ is the expected value.

Can anybody explain how the first approximation was achieved? It is just simply a cross multiplication in the limits? Thanks.

Answer 1

$$Af(x) = \lim_{t \downarrow 0} \frac{\mathbb{E}^x(f(X_t))-f(x)}{t}$$ implies that $$\lim_{t \downarrow 0} \mathbb{E}^x(f(X_t))-f(x) - tAf(x) = \lim_{t \downarrow 0} t \left[ \frac{\mathbb{E}^x(f(X_t))-f(x)}{t} - Af(x) \right] = 0$$ which tells you that $\mathbb{E}^x(f(X_t)) \approx f(x) + tAf(x)$ for small $t \geq 0$.