\documentclass[12pt]{article} \usepackage{amsmath,amsthm,amsfonts,amssymb,latexsym, hyperref} \begin{document} \begin{center} \textbf{Useful Limit Theorems.} \end{center} Reminder: In the following assume all functions have domain some open set and range a subset of the reals. Assume also that each formula is well-defined (e.g. $p$ is in the common domain where required.) [There are similar theorems for sequential limits.] \ Definition. Suppose $f: U \rightarrow \mathbb{R}$ is a function and $p$ is a number in the domain of $f$. Then $$\lim_{x \rightarrow p} f(x) = L$$ means that if $\epsilon >0$ then there exists a number $\delta$ so that \begin{eqnarray*} |f(x) - L| & < & \epsilon \end{eqnarray*} for all $x$ so that \begin{eqnarray*} 0 & < |x-p| < & \delta. \end{eqnarray*} \ In the following theorems assume all functions have the number $p$ in their domains. \ Theorem L1. $$\lim_{x \rightarrow p} (f(x) + g(x)) = \lim_{x \rightarrow p} f(x) + \lim_{x \rightarrow p} g(x).$$ \ Theorem L2. If $c$ is a constant then $$\lim_{x \rightarrow p} (cf(x)) = c \lim_{x \rightarrow p} f(x).$$ \ Theorem L3. $$\lim_{x \rightarrow p} (f(x) \cdot g(x)) = \lim_{x \rightarrow p} f(x) \cdot \lim_{x \rightarrow p} g(x).$$ \ Theorem L4. If $$\lim_{x \rightarrow p} f(x) \ne 0$$ then $$\lim_{x \rightarrow p} \frac 1 {f(x)} = \frac 1{\lim_{x \rightarrow p} f(x)} .$$ \end{document}