Let [latex]\epsilon >0.[/latex]

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

Choose [latex]\delta =\frac{\epsilon }{2}.[/latex]

The definition continues with “there exists a [latex]\delta >0.[/latex]” The phrase “there exists” in a mathematical statement is always a signal for a scavenger hunt. In other words, we must go and find [latex]\delta .[/latex] So, where exactly did [latex]\delta =\epsilon \text{/}2[/latex] come from? There are two basic approaches to tracking down [latex]\delta .[/latex] 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 [latex]|(2x+1)-3|<\epsilon ,[/latex] we begin by manipulating this expression: [latex]|(2x+1)-3|<\epsilon[/latex] is equivalent to [latex]|2x-2|<\epsilon ,[/latex] which in turn is equivalent to [latex]|2||x-1|<\epsilon .[/latex] Last, this is equivalent to [latex]|x-1|<\epsilon \text{/}2.[/latex] Thus, it would seem that [latex]\delta =\epsilon \text{/}2[/latex] is appropriate.

We may also find [latex]\delta[/latex] through geometric methods. (Figure) demonstrates how this is done.

Figure 2. This graph shows how we find [latex]\delta [/latex] geometrically.

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

Thus,

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

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

Let [latex]\epsilon >0.[/latex]

Choose [latex]\delta =\text{_______.}[/latex]

Assume [latex]0<|x-(-1)|<\delta .[/latex] (or equivalently, [latex]0<|x+1|<\delta \text{.)}[/latex]

Thus, [latex]|(4x+1)-(-3)|=|4x+4|=|4||x+1|<4\delta \text{_______}\epsilon .[/latex]

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

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

Let [latex]\epsilon >0.[/latex]

Choose [latex]\delta =\frac{\epsilon }{4}.[/latex]

Assume [latex]0<|x-(-1)|<\delta[/latex] (or equivalently, [latex]0<|x+1|<\delta \text{.)}[/latex]

Thus, [latex]|(4x+1)-(-3)|=|4x+4|=|4||x+1|<4\delta =4(\epsilon \text{/}4)=\epsilon .[/latex]

1. Let [latex]\epsilon >0.[/latex] The first part of the definition begins “For every [latex]\epsilon >0\text{,”}[/latex] so we must prove that whatever follows is true no matter what positive value of ε is chosen. By stating “Let [latex]\epsilon >0\text{,”}[/latex] we signal our intent to do so.
2. Without loss of generality, assume [latex]\epsilon \le 4.[/latex] Two questions present themselves: Why do we want [latex]\epsilon \le 4[/latex] and why is it okay to make this assumption? In answer to the first question: Later on, in the process of solving for [latex]\delta ,[/latex] we will discover that [latex]\delta[/latex] involves the quantity [latex]\sqrt{4-\epsilon }.[/latex] Consequently, we need [latex]\epsilon \le 4.[/latex] In answer to the second question: If we can find [latex]\delta >0[/latex] that “works” for [latex]\epsilon \le 4,[/latex] then it will “work” for any [latex]\epsilon >4[/latex] as well. Keep in mind that, although it is always okay to put an upper bound on ε, it is never okay to put a lower bound (other than zero) on ε.
3. Choose [latex]\delta =\text{min}\left\{2-\sqrt{4-\epsilon },\sqrt{4+\epsilon }-2\right\}.[/latex] (Figure) shows how we made this choice of [latex]\delta .[/latex]
   

   Figure 3. This graph shows how we find δ geometrically for a given ε for the proof in (Figure).

4. We must show: If [latex]0<|x-2|<\delta ,[/latex] then [latex]|{x}^{2}-4|<\epsilon ,[/latex] so we must begin by assuming
   [latex]0<|x-2|<\delta .[/latex]

   We don’t really need [latex]0<|x-2|[/latex] (in other words, [latex]x\ne 2)[/latex] for this proof. Since [latex]0<|x-2|<\delta ⇒|x-2|<\delta ,[/latex] it is okay to drop [latex]0<|x-2|.[/latex]

   [latex]|x-2|<\delta .[/latex]

   Hence,

   [latex]-\delta

   Recall that [latex]\delta =\text{min}\left\{2-\sqrt{4-\epsilon },\sqrt{4+\epsilon }-2\right\}.[/latex] Thus, [latex]\delta \ge 2-\sqrt{4-\epsilon }[/latex] and consequently [latex]-(2-\sqrt{4-\epsilon })\le -\delta .[/latex] We also use [latex]\delta \le \sqrt{4+\epsilon }-2[/latex] here. We might ask at this point: Why did we substitute [latex]2-\sqrt{4-\epsilon }[/latex] for [latex]\delta[/latex] on the left-hand side of the inequality and [latex]\sqrt{4+\epsilon }-2[/latex] on the right-hand side of the inequality? If we look at (Figure), we see that [latex]2-\sqrt{4-\epsilon }[/latex] corresponds to the distance on the left of 2 on the [latex]x[/latex]-axis and [latex]\sqrt{4+\epsilon }-2[/latex] corresponds to the distance on the right. Thus,

   [latex]-(2-\sqrt{4-\epsilon })\le -\delta

   We simplify the expression on the left:

   [latex]-2+\sqrt{4-\epsilon }

   Then, we add 2 to all parts of the inequality:

   [latex]\sqrt{4-\epsilon }

   We square all parts of the inequality. It is okay to do so, since all parts of the inequality are positive:

   [latex]4-\epsilon <{x}^{2}<4+\epsilon .[/latex]

   We subtract 4 from all parts of the inequality:

   [latex]-\epsilon <{x}^{2}-4<\epsilon .[/latex]

   Last,

   [latex]|{x}^{2}-4|<\epsilon .[/latex]
5. Therefore,
   [latex]\underset{x\to 2}{\text{lim}}{x}^{2}=4.[/latex]

Let’s use our outline from the Problem-Solving Strategy:

1. Let [latex]\epsilon >0.[/latex]
2. Choose [latex]\delta =\text{min}\left\{1,\epsilon \text{/}5\right\}.[/latex] This choice of [latex]\delta[/latex] may appear odd at first glance, but it was obtained by taking a look at our ultimate desired inequality: [latex]|({x}^{2}-2x+3)-6|<\epsilon .[/latex] This inequality is equivalent to [latex]|x+1|·|x-3|<\epsilon .[/latex] At this point, the temptation simply to choose [latex]\delta =\frac{\epsilon }{x-3}[/latex] is very strong. Unfortunately, our choice of [latex]\delta[/latex] must depend on ε only and no other variable. If we can replace [latex]|x-3|[/latex] by a numerical value, our problem can be resolved. This is the place where assuming [latex]\delta \le 1[/latex] comes into play. The choice of [latex]\delta \le 1[/latex] here is arbitrary. We could have just as easily used any other positive number. In some proofs, greater care in this choice may be necessary. Now, since [latex]\delta \le 1[/latex] and [latex]|x+1|<\delta \le 1,[/latex] we are able to show that [latex]|x-3|<5.[/latex] Consequently, [latex]|x+1|·|x-3|<|x+1|·5.[/latex] At this point we realize that we also need [latex]\delta \le \epsilon \text{/}5.[/latex] Thus, we choose [latex]\delta =\text{min}\left\{1,\epsilon \text{/}5\right\}.[/latex]
3. Assume [latex]0<|x+1|<\delta .[/latex] Thus,
   [latex]|x+1|<1\text{ and }|x+1|<\frac{\epsilon }{5}.[/latex]

   Since [latex]|x+1|<1,[/latex] we may conclude that [latex]-1[latex]|({x}^{2}-2x+3)-6|=|x+1|·|x-3|<\frac{\epsilon }{5}·5=\epsilon .[/latex]