Art of Problem Solving

Art of Problem Solving

AoPS Online

User:Temperal/The Problem Solver's Resource7: Difference between revisions

Latest revision as of 09:59, 19 June 2025

Introduction	Other Tips and Tricks	Methods of Proof	You are currently viewing page 7.

Limits

This section covers limits and some other precalculus topics.

Definition

$\lim_{x\to n}f(x)$ is the value that $f(x)$ approaches as $x$ approaches $n$ .

$\lim_{x\uparrow n}f(x)$ is the value that $f(x)$ approaches as $x$ approaches $n$ from values of $x$ less than $n$ .

$\lim_{x\downarrow n}f(x)$ is the value that $f(x)$ approaches as $x$ approaches $n$ from values of $x$ more than $n$ .

If $\lim_{x\to n}f(x)=f(n)$ , then $f(x)$ is said to be continuous in $n$ .

Properties

Let $f$ and $g$ be real functions. Then:

$\lim(f+g)(x)=\lim f(x)+\lim g(x)$
$\lim(f-g)(x)=\lim f(x)-\lim g(x)$
$\lim(f\cdot g)(x)=\lim f(x)\cdot\lim g(x)$
$\lim\left(\frac{f}{g}\right)(x)=\frac{\lim f(x)}{\lim g(x)}$

Squeeze Play Theorem (or Sandwich Theorem)

Suppose $f(x)$ is between $g(x)$ and $h(x)$ for all $x$ in the neighborhood of $S$ . If $g$ and $h$ approach some common limit L as $x$ approaches $S$ , then $\lim_{x\to S}f(x)=L$ .

Diverging-Converging Theorem

A series $\sum_{i=0}^{\infty}S_i$ converges iff $\lim S_i=0$ .

Focus Theorem

The statement $\lim_{x\to n}f(x)=L$ is equivalent to: given a positive number $\epsilon$ , there is a positive number $\gamma$ such that $0<|x-n|<\gamma\Rightarrow |f(x)-L|<\epsilon$ .

Back to page 6 | Continue to page 8

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource7&oldid=251558"

© 2025 AoPS Incorporated