Q. 69 Prove that limx鈫�2(7-x)=5, with... [FREE SOLUTION]

Chapter 1: Q. 69 (page 99)

Prove that $\lim_{x 鈫� 2} (7 - x) = 5$ , with these steps:
(a) What is the $未 - 蔚$ statement that must be shown to prove that $\lim_{x 鈫� 2} (7 - x) = 5$ ?
(b) Argue that $x 鈭� (2 - 未, 2) 鈭� (2, 2 + 未)$ if and only if $- 未 < x - 2 < 未$ . Then use algebra to show that this means that $0 < |x - 2| < 未$ .
(c) Argue that $7 - x 鈭� (5 - 蔚, 5 + 蔚)$ if and only if $- 蔚 < 2 - x < 蔚$ . Then use algebra to show that this means that $|x - 2| < 蔚$ .
(d) Given any particular $蔚 > 0$ , what value of $未$ would guarantee that if $0 < |x - 2| < 未$ , then $|x - 2| < 蔚$ ? Your answer will depend on $蔚$ .
(e) Put the previous four parts together to prove the limit statement.

Short Answer

Expert verified

$Part (a) For all 蔚 > 0, there exists 未 > 0 such that if x 鈭� (2 鈭� 未, 2) 鈭� (2, 2 + 未), then 7 鈭� x 鈭� (5 鈭� 蔚, 5 + 蔚) . Part (b) 2 鈭� 未 < x < 2 + 未 and x 鈮� 2 means that 鈭� 未 < x 鈭� 2 < 未 and x 鈮� 2 . This means that | x 鈭� 2 | < 未 and x 鈭� 2 鈮� 0, and thus 0 < | x 鈭� 2 | < 未 . Part (c) 5 鈭� 蔚 < 7 鈭� x < 5 + 蔚 means that 鈭� 蔚 < 2 鈭� x < 蔚, and thus that | 2 鈭� x | < 蔚, which is equivalent to | x 鈭� 2 | < 蔚 . Part (d) For a given 蔚 > 0, choose 未 = 蔚 . Part (e) Given any 蔚 > 0, let 未 = 蔚 . If x 鈭� (2 鈭� 未, 2) 鈭� (2, 2 + 未), then 0 < | x 鈭� 2 | < 未 by part (b) . Thus by part (d), | x 鈭� 2 | < 蔚, and therefore 7 鈭� x 鈭� (5 鈭� 蔚, 5 + 蔚) by part (c) .$

Step by step solution

Part (a) Step 1. The δ-ε statement

$For all 蔚 > 0, there exists 未 > 0 such that if x 鈭� (2 鈭� 未, 2) 鈭� (2, 2 + 未), then 7 鈭� x 鈭� (5 鈭� 蔚, 5 + 蔚) .$

Part (b) Step 1. Argument regarding δ

$Now, 2 鈭� 未 < x < 2 + 未 and x 鈮� 2 means that 鈭� 未 < x 鈭� 2 < 未 and x 鈮� 2 . This means that | x 鈭� 2 | < 未 and x 鈭� 2 鈮� 0, and thus 0 < | x 鈭� 2 | < 未 .$

Part (c) Step 1. Argument regarding ε

$Again, 5 鈭� 蔚 < 7 鈭� x < 5 + 蔚 means that 鈭� 蔚 < 2 鈭� x < 蔚, and thus that | 2 鈭� x | < 蔚, which is equivalent to | x 鈭� 2 | < 蔚 .$

Part (d) Step 1. Relationship between δ-ε

$For a given 蔚 > 0, choose 未 = 蔚 .$

Part (e) Step 1. Proof of the limit statement

$Given any 蔚 > 0, let 未 = 蔚 . If x 鈭� (2 鈭� 未, 2) 鈭� (2, 2 + 未), then 0 < | x 鈭� 2 | < 未 by part (b) . Thus by part (d), | x 鈭� 2 | < 蔚, and therefore 7 鈭� x 鈭� (5 鈭� 蔚, 5 + 蔚) by part (c) .$

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Short Answer

Step by step solution

Part (a) Step 1. The δ-ε statement

Part (b) Step 1. Argument regarding δ

Part (c) Step 1. Argument regarding ε

Part (d) Step 1. Relationship between δ-ε

Part (e) Step 1. Proof of the limit statement

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