Q. 0 - Calculus

Question

Read the section and make your own summary of the material.

Step-by-Step Solution

Answer

Limit $\underset{x\to c}{\lim }f\left(x\right)=L$state that the $\delta >0$exists for all  $\epsilon >0$ such that,

If $0<|x-c|<\delta$, then $\left|f\right(x)-L|<\epsilon .$

The theorem of Equivalence of Geometric and Algebraic Definitions of the Limit state,

$$\begin{gathered} \left(a\right) x\in (c-\delta ,c)\cup (c,c+\delta ) \mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if} 0<|x-c|<\delta . \\ \left(b\right) f\left(x\right)\in (L-\epsilon ,L+\epsilon ) \mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if} \left|f\right(x)-L|<\epsilon . \end{gathered}$$

Step 1. Given information.

Subchapter of delta epsilon proofs.

Step 2. Summary of chapter.

The algebraic definition of Limit is following.

Limit $\underset{x\to c}{\lim }f\left(x\right)=L$state that the $\delta >0$exists for all $\epsilon >0$ such that,

If $0<|x-c|<\delta$, then $\left|f\right(x)-L|<\epsilon .$

The theorem of Equivalence of Geometric and Algebraic Definitions of the Limit state the following statements.

$$\begin{gathered} \left(a\right) x\in (c-\delta ,c)\cup (c,c+\delta ) \mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if} 0<|x-c|<\delta . \\ \left(b\right) f\left(x\right)\in (L-\epsilon ,L+\epsilon ) \mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if} \left|f\right(x)-L|<\epsilon . \end{gathered}$$