Chapter 2 | Limits
199
Proving a Statement about the Limit of a Specific Function (Geometric Approach)
x 2 =4.
Prove that lim x →2
Solution 1. Let ε >0. The first part of the definition begins “For every ε >0,” so we must prove that whatever follows is true no matter what positive value of ε is chosen. By stating “Let ε >0,” we signal our intent to do so. 2. Without loss of generality, assume ε ≤4. Two questions present themselves: Why do we want ε ≤4 and why is it okay to make this assumption? In answer to the first question: Later on, in the process of solving for δ , we will discover that δ involves the quantity 4− ε . Consequently, we need ε ≤4. In answer to the second question: If we can find δ >0 that “works” for ε ≤4, then it will “work” for any ε >4 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 δ =min ⎧ ⎩ ⎨ 2− 4− ε , 4+ ε −2 ⎫ ⎭ ⎬ . Figure 2.41 shows how we made this choice of δ .
Figure 2.41 This graph shows how we find δ geometrically for a given ε for the proof in Example 2.41 . 4. We must show: If 0< | x −2| < δ , then | x 2 −4| < ε , so we must begin by assuming 0< | x −2| < δ . We don’t really need 0< | x −2| (in other words, x ≠2) for this proof. Since 0< | x −2| < δ ⇒ | x −2| < δ , it is okay to drop 0< | x −2|. | x −2| < δ .
Hence,
− δ < x −2< δ .
δ =min ⎧ ⎩
⎨ 2− 4− ε , 4+ ε −2 ⎫ ⎭ ⎬ .
δ ≤2− 4− ε
Recall
that
Thus,
and consequently
− ⎛
⎝ 2− 4− ε ⎞
⎠ ≤ − δ . We also use δ ≤ 4+ ε −2 here. We might ask at this point: Why did we substitute 2− 4− ε for δ on the left-hand side of the inequality and 4+ ε −2 on the right-hand side of the inequality? If we look at Figure 2.41 , we see that 2− 4− ε corresponds to the distance on
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