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Chapter 2 | Limits
Example 2.40 Proving a Statement about a Limit
(4 x +1) =−3 by filling in the blanks.
Complete the proof that lim x →−1
Let _____. Choose δ =_______. Assume 0< | x − _______| < δ . Thus, |________ − ________| = _____________________________________ ε . Solution We begin by filling in the blanks where the choices are specified by the definition. Thus, we have Let ε >0. Choose δ =_______.
Assume 0< | x −(−1) | < δ . (or equivalently, 0< | x +1 | < δ .) Thus, | (4 x +1)−(−3) | = | 4 x +4 | = |4| | x +1 | <4 δ _______ ε . Focusing on the final line of the proof, we see that we should choose δ = ε 4 . We now complete the final write-up of the proof: Let ε >0. Choose δ = ε 4 . Assume 0< | x −(−1) | < δ (or equivalently, 0< | x +1 | < δ .) Thus, | (4 x +1)−(−3) | = | 4 x +4 | = |4| | x +1 | <4 δ =4( ε /4) = ε .
(3 x −2) =4 by filling in the blanks.
Complete the proof that lim x →2
2.27
Let _______. Choose δ =_______. Assume 0< | x − ____| < ____. Thus, |_______ − ____| = ______________________________ ε . Therefore, lim x →2 (3 x −2) =4.
In Example 2.39 and Example 2.40 , the proofs were fairly straightforward, since the functions with which we were working were linear. In Example 2.41 , we see how to modify the proof to accommodate a nonlinear function. Example 2.41
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