If both sides exist, it can be proved that:
$$ \begin{aligned} &\mathop {\lim }\limits_{x \to a} (f(x) + g(x)) = \mathop {\lim }\limits_{x \to a} f(x) + \mathop {\lim }\limits_{x \to a} g(x) \\ \\ &\mathop {\lim }\limits_{x \to a} c \cdot f(x) = c \cdot \mathop {\lim }\limits_{x \to a} g(x) \\ \\ &\mathop {\lim }\limits_{x \to a} (f(x) \cdot g(x)) = \mathop {\lim }\limits_{x \to a} f(x) \cdot \mathop {\lim }\limits_{x \to a} g(x) \\ \\ &\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}} \ (\text{if} \ \mathop {\lim }\limits_{x \to a} g(x) \ne 0) \\ \\ &\mathop {\lim }\limits_{x \to a} f{(x)^p} = {\left( {\mathop {\lim }\limits_{x \to a} f(x)} \right)^p} \\ \\ &\mathop {\lim }\limits_{x \to a} \left| {f(x)} \right| = \left| {\mathop {\lim }\limits_{x \to a} f(x)} \right| \end{aligned} $$