Basic Operation Rules

1. Since $\lim f(x) = A$, for any $\varepsilon > 0$, there exists $\delta_1 > 0$ such that when $0 < \vert x - x_0 \vert < \delta_1$, we have $\vert f(x) - A \vert < \frac{\varepsilon}{2}$

2. Since $\lim g(x) = B$, for any $\varepsilon > 0$, there exists $\delta_2 > 0$ such that when $0 < \vert x - x_0 \vert < \delta_2$, we have $\vert g(x) - B \vert < \frac{\varepsilon}{2}$

3. Let $\delta = \min\{\delta_1, \delta_2\}$; then when $0 < \vert x - x_0 \vert < \delta$: $\vert f(x) + g(x) - (A + B) \vert \leq \vert f(x) - A \vert + \vert g(x) - B \vert < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$

4. Therefore $\lim [f(x) + g(x)] = A + B$

Idea of the proof:

1. Use the identity: $f(x) \cdot g(x) - A \cdot B = [f(x) - A] \cdot g(x) + A \cdot [g(x) - B]$

2. Since $\lim g(x) = B$, $g(x)$ is bounded in some neighborhood of $x_0$

3. Use the property of infinitesimals: the product of a bounded function and an infinitesimal is still an infinitesimal

4. Conclude $\lim [f(x) \cdot g(x)] = A \cdot B$

Generalization: $\lim [c \cdot f(x)] = c \cdot A$ (where $c$ is a constant)

Idea of the proof:

1. First prove $\lim \frac{1}{g(x)} = \frac{1}{B}$

2. Since $B \neq 0$, $g(x)$ is nonzero in some neighborhood of $x_0$

3. Use the multiplication rule: $\lim \frac{f(x)}{g(x)} = \lim f(x) \cdot \lim \frac{1}{g(x)} = A \cdot \frac{1}{B} = \frac{A}{B}$

Generalization: $\lim [f(x)]^{g(x)} = A^B$ (subject to certain conditions)