Q. 68 Prove聽that聽lim聽3x鈫�1x=3,聽wi... [FREE SOLUTION]

Chapter 1: Q. 68 (page 99)

$Prove that \underset{x 鈫� 1}{\lim 3} x = 3, with these steps : (a) What is the 未鈥� 蔚 statement that must be shown to prove that \underset{x 鈫� 1}{\lim 3} x = 3 ? (b) Argue that x 鈭� (1 鈭� 未, 1) 鈭� (1, 1 + 未) if and only if 鈭� 未 < x 鈭� 1 < 未, w i t h x 鈮� 1 . Then use algebra to show that this means that 0 < | x 鈭� 1 | < 未 . (c) Argue that 3 x 鈭� (3 鈭� 蔚, 3 + 蔚) if and only if 鈭� 蔚 < 3 (x 鈭� 1) < 蔚 . Then use algebra to show that this means that 3 | x 鈭� 1 | < 蔚 . (d) Given any particular 蔚 > 0, what value of 未 would guarantee that if 0 < | x 鈭� 1 | < 未, then 3 | x 鈭� 1 | < 蔚 ? Your answer will depend on 蔚 . (e) Put the previous four parts together to prove the limit statement .$

Short Answer

Expert verified

$Part (a) For all 蔚 > 0, there exists 未 > 0 such that if x 鈭� (1 鈭� 未, 1) 鈭� (1, 1 + 未), then 3 x 鈭� (3 鈭� 蔚, 3 + 蔚) . Part (b) 1 鈭� 未 < x < 1 + 未 and x 鈮� 1 means that 鈭� 1 < x 鈭� 1 < 1 and x 鈮� 1 . This means that | x 鈭� 1 | < 未 and x 鈭� 1 鈮� 0, and thus 0 < | x 鈭� 1 | < 未 . Part (c) 3 鈭� 蔚 < 3 x < 3 + 蔚 means that 鈭� 蔚 < 3 x - 3 < 蔚, and thus that | 3 x - 3 | < 蔚, which is equivalent to 3 | x 鈭� 1 | < 蔚 . Part (d) For a given 蔚 > 0, choose 未 = 蔚 . Part (e) Given any 蔚 > 0, let 未 = 蔚 . If x 鈭� (1 鈭� 未, 1) 鈭� (1, 1 + 未), then 0 < | x 鈭� 1 | < 未 by part (b) . Thus by part (d), 3 | x 鈭� 1 | < 蔚, and therefore 3 x 鈭� (3 鈭� 蔚, 3 + 蔚) by part (c) .$

Step by step solution

Part (a) Step 1. The δ-ε statement

$For all 蔚 > 0, there exists 未 > 0 such that if x 鈭� (1 鈭� 未, 1) 鈭� (1, 1 + 未), then 3 x 鈭� (3 鈭� 蔚, 3 + 蔚) .$

Part (b) Step 1. Argument regarding δ

$Now, 1 鈭� 未 < x < 1 + 未 and x 鈮� 1 means that 鈭� 1 < x 鈭� 1 < 1 and x 鈮� 1 . This means that | x 鈭� 1 | < 未 and x 鈭� 1 鈮� 0, and thus 0 < | x 鈭� 1 | < 未 .$

Part (c) Step 1. Argument regarding ε

$Again, 3 鈭� 蔚 < 3 x < 3 + 蔚 means that 鈭� 蔚 < 3 x - 3 < 蔚, and thus that | 3 x - 3 | < 蔚, which is equivalent to 3 | x 鈭� 1 | < 蔚 .$

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

$Given any 蔚 > 0, let 未 = 蔚 . If x 鈭� (1 鈭� 未, 1) 鈭� (1, 1 + 未), then 0 < | x 鈭� 1 | < 未 by part (b) . Thus by part (d), 3 | x 鈭� 1 | < 蔚, and therefore 3 x 鈭� (3 鈭� 蔚, 3 + 蔚) by part (c) .$

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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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