2.5 The precise definition of a limit  (Page 2/8)

Let $\epsilon >0.$

The first part of the definition begins “For every $\epsilon >0\text{.”}$ This means we must prove that whatever follows is true no matter what positive value of ε is chosen. By stating “Let $\epsilon >0\text{,”}$ we signal our intent to do so.

Choose $\delta =\frac{\epsilon }{2}.$

The definition continues with “there exists a $\delta >0.$ ” The phrase “there exists” in a mathematical statement is always a signal for a scavenger hunt. In other words, we must go and find $\delta .$ So, where exactly did $\delta =\epsilon \text{/}2$ come from? There are two basic approaches to tracking down $\delta .$ One method is purely algebraic and the other is geometric.

We begin by tackling the problem from an algebraic point of view. Since ultimately we want $\left|\left(2x+1\right)-3\right|<\epsilon ,$ we begin by manipulating this expression: $\left|\left(2x+1\right)-3\right|<\epsilon$ is equivalent to $\left|2x-2\right|<\epsilon ,$ which in turn is equivalent to $\left|2\right|\left|x-1\right|<\epsilon .$ Last, this is equivalent to $\left|x-1\right|<\epsilon \text{/}2.$ Thus, it would seem that $\delta =\epsilon \text{/}2$ is appropriate.

We may also find $\delta$ through geometric methods. [link] demonstrates how this is done.

Assume $0<\left|x-1\right|<\delta .$ When $\delta$ has been chosen, our goal is to show that if $0<\left|x-1\right|<\delta ,$ then $\left|\left(2x+1\right)-3\right|<\epsilon .$ To prove any statement of the form “If this, then that,” we begin by assuming “this” and trying to get “that.”

Thus,

$\begin{array}{ccccc}\left|\left(2x+1\right)-3\right|\hfill & =\left|2x-2\right|\hfill & & & \text{property of absolute value}\hfill \\ & =\left|2\left(x-1\right)\right|\hfill \\ & =\left|2\right|\left|x-1\right|\hfill & & & \left|2\right|=2\hfill \\ & =2\left|x-1\right|\hfill & & & \\ & <2·\delta \hfill & & & \text{here’s where we use the assumption that}0<\left|x-1\right|<\delta \hfill \\ & =2·\frac{\epsilon }{2}=\epsilon \hfill & & & \text{here’s where we use our choice of}\delta =\epsilon \text{/}2\hfill \end{array}$

We begin by filling in the blanks where the choices are specified by the definition. Thus, we have

Let $\epsilon >0.$

Choose $\delta =\text{\_\_\_\_\_\_\_.}$

Assume $0<\left|x-\left(\mathrm{-1}\right)\right|<\delta .$ (or equivalently, $0<\left|x+1\right|<\delta \text{.)}$

Thus, $\left|\left(4x+1\right)-\left(\mathrm{-3}\right)\right|=\left|4x+4\right|=\left|4\right|\left|x+1\right|<4\delta \text{\_\_\_\_\_\_\_}\epsilon .$

Focusing on the final line of the proof, we see that we should choose $\delta =\frac{\epsilon }{4}.$

We now complete the final write-up of the proof:

Let $\epsilon >0.$

Choose $\delta =\frac{\epsilon }{4}.$

Assume $0<\left|x-\left(\mathrm{-1}\right)\right|<\delta$ (or equivalently, $0<\left|x+1\right|<\delta \text{.)}$

Thus, $\left|\left(4x+1\right)-\left(\mathrm{-3}\right)\right|=\left|4x+4\right|=\left|4\right|\left|x+1\right|<4\delta =4\left(\epsilon \text{/}4\right)=\epsilon .$