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

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==<span style="font-size:20px; color: blue;">Limits</span>==
 
==<span style="font-size:20px; color: blue;">Limits</span>==
 
This section covers limits and some other precalculus topics.
 
This section covers limits and some other precalculus topics.
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*If <math>\lim_{x\to n}f(x)=f(n)</math>, then <math>f(x)</math> is said to be continuous in <math>n</math>.
 
*If <math>\lim_{x\to n}f(x)=f(n)</math>, then <math>f(x)</math> is said to be continuous in <math>n</math>.
  
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===Theorems and Properties===
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===Properties===
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The statement <math>\lim_{x\to n}f(x)=L</math> is equivalent to: given a positive number <math>\epsilon</math>, there is a positive number <math>\gamma</math> such that <math>0<|x-n|<\gamma\Rightarrow |f(x)-L|<\epsilon</math>.
 
  
 
Let <math>f</math> and <math>g</math> be real functions. Then:
 
Let <math>f</math> and <math>g</math> be real functions. Then:
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*<math>\lim\left(\frac{f}{g}\right)(x)=\frac{\lim f(x)}{\lim g(x)}</math>
 
*<math>\lim\left(\frac{f}{g}\right)(x)=\frac{\lim f(x)}{\lim g(x)}</math>
  
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===Squeeze Play Theorem (or Sandwich Theorem)===
 
Suppose <math>f(x)</math> is between <math>g(x)</math> and <math>h(x)</math> for all <math>x</math> in the neighborhood of <math>S</math>. If <math>g</math> and <math>h</math> approach some common limit L as <math>x</math> approaches <math>S</math>, then <math>\lim_{x\to S}f(x)=L</math>.
 
Suppose <math>f(x)</math> is between <math>g(x)</math> and <math>h(x)</math> for all <math>x</math> in the neighborhood of <math>S</math>. If <math>g</math> and <math>h</math> approach some common limit L as <math>x</math> approaches <math>S</math>, then <math>\lim_{x\to S}f(x)=L</math>.
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===Diverging-Converging Theorem===
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A series <math>\sum_{i=0}^{\infty}S_i</math> converges iff <math>\lim S_i=0</math>.
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===Focus Theorem===
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The statement <math>\lim_{x\to n}f(x)=L</math> is equivalent to: given a positive number <math>\epsilon</math>, there is a positive number <math>\gamma</math> such that <math>0<|x-n|<\gamma\Rightarrow |f(x)-L|<\epsilon</math>.
  
  
 
[[User:Temperal/The Problem Solver's Resource6|Back to page 6]] | [[User:Temperal/The Problem Solver's Resource8|Continue to page 8]]
 
[[User:Temperal/The Problem Solver's Resource6|Back to page 6]] | [[User:Temperal/The Problem Solver's Resource8|Continue to page 8]]
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Latest revision as of 18:19, 10 January 2009

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 $0<|x-n|<\gamma\Rightarrow |f(x)-L|<\epsilon$.


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