User:Derserrs/Calculus 101

Calculus 101 (under construction)
Introduction

Calculus is a branch of mathematics that basically deals with limits. It is built primarily on two different ideas: differential calculus and integral calculus. The former is based on the slope of a function's graph, and the latter on the area below a function's curve.

To be able to have a knowledge of these two areas, it is necessary to first have an understanding of the concept limit.

Limits of Sequences
"Limit" is the term used to describe the behavior of a sequence (or function) as it gets close to a given point. For example, as n gets bigger and bigger, the sequence $$A_n=\frac{1}{n}$$ gets closer and closer to 0. Hence, we say that the limit of $$A_n$$, as n approaches infinity, is 0.

$$\lim_{n\to \infty}\frac{1}{n}=0$$

Here's the formal definition:

$$\lim_{n\to \infty}A_n=L$$ $$\Rightarrow$$       If given an $$\epsilon>0$$, there exists an $$N$$ where if $$n\ge N$$, then $$|A_n-L|<\epsilon$$

We already know one limit:

$$\lim_{n\to \infty}\frac{1}{n}=0$$

Proof:

Given an $$\epsilon>0$$, we choose an $$N$$ so that $$N>\frac{1}{\epsilon}$$

Using the definition: $$|\frac{1}{n}-0|<\epsilon$$ $$\Rightarrow$$    $$\frac{1}{n}<\epsilon$$   $$\Rightarrow$$   $$\frac{1}{\epsilon}0$$

Proof:

$$|\frac{1}{n^p}-0|<\epsilon \Rightarrow \frac{1}{n^p}<\epsilon  \Rightarrow  \frac{1}{\epsilon}\left(\frac{1}{\epsilon}\right)^{\frac{1}{p}}$$. By definition, if we can find an $$n>N$$, then $$n>\left(\frac{1}{\epsilon}\right)^{\frac{1}{p}} \Rightarrow n^p>\frac{1}{\epsilon} \Rightarrow \epsilon>\frac{1}{n^p} \Rightarrow \epsilon>|\frac{1}{n^p}-0|$$

We're back where we started and finished with the proof.

Using the definition of limit, the following theorems can easily be proven:


 * 1) If $$A_n=C$$ (constant), then $$\lim_{n\to \infty}A_n=\lim_{n\to \infty}C=C$$
 * 2) If $$\lim_{n\to \infty}A_n=X$$ and $$\lim_{n\to \infty}B_n=Y$$, then:
 * a)$$\lim_{n\to \infty}(A_n+B_n)=X+Y$$
 * b)$$\lim_{n\to \infty}(A_nB_n)=XY$$
 * c)$$\lim_{n\to \infty}\frac{A_n}{B_n}=\frac{X}{Y}$$ if  $$Y\neq0$$