Chapter 2 | Limits
197
In this part of the proof, we started with | (2 x +1)−3 | and used our assumption 0< | x −1| < δ in a key part of the chain of inequalities to get | (2 x +1)−3 | to be less than ε . We could just as easily have manipulated the assumed inequality 0< | x −1| < δ to arrive at | (2 x +1)−3 | < ε as follows: 0< | x −1| < δ ⇒ | x −1| < δ ⇒ − δ < x −1< δ ⇒ − ε 2 < x −1< ε 2
⇒ − ε <2 x −2< ε ⇒ − ε <2 x −2< ε ⇒ |2 x −2| < ε ⇒ | (2 x +1)−3 | < ε .
(2 x +1) =3. (Having completed the proof, we state what we have accomplished.)
Therefore, lim x →1
After removing all the remarks, here is a final version of the proof: Let ε >0. Choose δ = ε /2. Assume 0< | x −1| < δ . Thus, | (2 x +1)−3 | = |2 x −2|
= | 2( x −1) | = |2|| x −1| =2| x −1|
<2· δ =2· ε 2 = ε .
(2 x +1) =3.
Therefore, lim x →1
The following Problem-Solving Strategy summarizes the type of proof we worked out in Example 2.39 .
x → a f ( x ) = L
for a Specific Function f ( x )
Problem-Solving Strategy: Proving That lim
1. Let’s begin the proof with the following statement: Let ε >0. 2. Next, we need to obtain a value for δ . After we have obtained this value, we make the following statement, filling in the blank with our choice of δ : Choose δ =_______. 3. The next statement in the proof should be (at this point, we fill in our given value for a ): Assume 0< | x − a | < δ . 4. Next, based on this assumption, we need to show that | f ( x )− L | < ε , where f ( x ) and L are our function f ( x ) and our limit L . At some point, we need to use 0< | x − a | < δ . 5. We conclude our proof with the statement: Therefore, lim x → a f ( x ) = L .
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