200
Chapter 2 | Limits
the left of 2 on the x -axis and 4+ ε −2 corresponds to the distance on the right. Thus, − ⎛ ⎝ 2− 4− ε ⎞ ⎠ ≤ − δ < x −2< δ ≤ 4+ ε −2.
We simplify the expression on the left:
−2+ 4− ε < x −2< 4+ ε −2.
Then, we add 2 to all parts of the inequality:
4− ε < x < 4+ ε .
We square all parts of the inequality. It is okay to do so, since all parts of the inequality are positive: 4− ε < x 2 <4+ ε .
We subtract 4 from all parts of the inequality:
− ε < x 2 −4< ε .
Last,
| x 2 −4| < ε .
5. Therefore,
x 2 =4.
lim x →2
Find δ corresponding to ε >0 for a proof that lim x →9
x =3.
2.28
The geometric approach to proving that the limit of a function takes on a specific value works quite well for some functions. Also, the insight into the formal definition of the limit that this method provides is invaluable. However, we may also approach limit proofs from a purely algebraic point of view. In many cases, an algebraic approach may not only provide us with additional insight into the definition, it may prove to be simpler as well. Furthermore, an algebraic approach is the primary tool used in proofs of statements about limits. For Example 2.42 , we take on a purely algebraic approach. Example 2.42 Proving a Statement about the Limit of a Specific Function (Algebraic Approach)
Prove that lim x →−1 ⎛
⎝ x 2 −2 x +3 ⎞
⎠ =6.
Solution Let’s use our outline from the Problem-Solving Strategy: 1. Let ε >0. 2. Choose δ =min{1, ε /5}. This choice of δ may appear odd at first glance, but it was obtained by
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