Solution

Let [latex]\epsilon \symbol{"3E} 0[/latex].

The first part of the definition begins “For every [latex]\epsilon \symbol{"3E} 0[/latex].” This means we must prove that whatever follows is true no matter what positive value of [latex]\epsilon[/latex] is chosen. By stating “Let [latex]\epsilon \symbol{"3E} 0[/latex],” we signal our intent to do so.

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

The definition continues with “there exists a [latex]\delta \symbol{"3E} 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/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/2[/latex]. Thus, it would seem that [latex]\delta =\epsilon/2[/latex] is appropriate.

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

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}{lllll}|(2x+1)-3| & =|2x-2| & & & \\ & =|2(x-1)| \\ & =|2||x-1| & & & \text{property of absolute values:} \, |ab|=|a||b| \\ & =2|x-1| & & & \\ & <2 \cdot \delta & & & \text{here’s where we use the assumption that} \, 0<|x-1|<\delta \\ & =2 \cdot \frac{\epsilon}{2}=\epsilon & & & \text{here’s where we use our choice of} \, \delta =\epsilon/2 \end{array}[/latex]

Solution

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

Let [latex]\epsilon \symbol{"3E} 0[/latex].

Choose [latex]\delta =[/latex] _______. (Leave this one blank for now — we’ll choose [latex]\delta[/latex] later)

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

Thus, [latex]|(4x+1)-(-3)|=|4x+4|=|4||x+1|<4\delta = \_\_\_\_\_\_\_\_\_\_\_\_ = \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 \symbol{"3E} 0[/latex].

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

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

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

Solution

1. Let [latex]\epsilon \symbol{"3E} 0[/latex]. The first part of the definition begins “For every [latex]\epsilon \symbol{"3E} 0[/latex],” so we must prove that whatever follows is true no matter what positive value of [latex]\epsilon[/latex] is chosen. By stating “Let [latex]\epsilon \symbol{"3E} 0[/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 \symbol{"3E} 0[/latex] that “works” for [latex]\epsilon \le 4[/latex], then it will “work” for any [latex]\epsilon \symbol{"3E} 4[/latex] as well. Keep in mind that, although it is always okay to put an upper bound on [latex]\epsilon[/latex], it is never okay to put a lower bound (other than zero) on [latex]\epsilon[/latex].
3. Choose [latex]\delta =\text{min}\{2-\sqrt{4-\epsilon},\sqrt{4+\epsilon}-2\}[/latex].  (Figure) shows how we made this choice of [latex]\delta[/latex].

   

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 \implies |x-2|<\delta[/latex], it is okay to drop [latex]0<|x-2|[/latex].

   So, [latex]|x-2|<\delta[/latex], which implies [latex]-\delta

   Recall that [latex]\delta =\text{min}\{2-\sqrt{4-\epsilon},\sqrt{4+\epsilon}-2\}[/latex]. Thus, [latex]\delta \le 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

   We subtract 4 from all parts of the inequality:

   [latex]-\epsilon

   Last,

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

Solution

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

1. Let [latex]\epsilon \symbol{"3E} 0[/latex].
2. Choose [latex]\delta =\text{min}\{1,\epsilon/5\}[/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|\cdot |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 [latex]\epsilon[/latex] 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| \cdot |x-3|<|x+1| \cdot 5[/latex]. At this point we realize that we also need [latex]\delta \le \epsilon/5[/latex]. Thus, we choose [latex]\delta =\text{min}\{1,\epsilon/5\}[/latex].
3. Assume [latex]0<|x+1|<\delta[/latex]. Thus,
   [latex]|x+1|<1[/latex] and [latex]|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| \cdot |x-3|<\frac{\epsilon}{5} \cdot 5=\epsilon[/latex].