Explanation Given: o Adapt the argument in Example 3 o Limit: lim x → c x − 1 = 1 c for all c ≠ 0 lim x → c f ( x ) = L is true for any function f ( x ) which is defined on an open interval around x = c , and L is a real number. That means for every number ε > 0 there exists some number δ > 0 such that if 0 < | x − c | < δ there exists | f ( x ) − L | < ε . Compare the given limit lim x → c x − 1 = 1 c with general definition of limit lim x → c f ( x ) = L and write the corresponding values of f ( x ) , c , a n d L . By the formal definition of limit, define the values of ε a n d δ . By the definition, if we prove that for any value of 0 < | x − c | < δ there exists | f ( x ) − L | < ε then the limit is vigorously true. Calculation: Description: Step 1 : Write the given function and consider this as f ( x ) . f ( x ) = x − 1 . This can also be written as f ( x ) = 1 x . Step 2 : Compare the given limit lim x → c x − 1 = 1 c with general definition of limit lim x → c f ( x ) = L and write the corresponding values of f ( x ) , c , a n d L . f ( x ) = 1 x c = c L = 1 c Step 3 : By the formal definition of limit, define the values of ε a n d δ . ε > 0 δ > 0 Step 4 : By the definition, if we prove that for any value of 0 < | x − c | < δ there exists | f ( x ) − L | < ε then the given limit exists and equals the given value which is vigorously true. For the given problem, 0 < | x − c | < δ becomes as 0 < | x − c | < δ . For the given problem, | f ( x ) − L | < ε becomes as | 1 x − 1 c | < ε . Step 5 : Write the value of | 1 x − 1 c | in expanded form. No need to bother about the numerator | x − c | . We have to consider the denominators | x | , | c |

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Question

Chapter 2.9, Problem 18E

To determine

To Prove:

Rigorously that the given limit existsfor all c except c≠0

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Chapter 2.9, Problem 17E

Chapter 2.9, Problem 19E

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Show that exp(2²)| ≤ exp(2²) < for all = E C.

i.) lim x→1 x² + x -2/x - 1; a. equals 3 b. equals 5 c. equals 4 d. none of the above/other answers are correct ii.) If f(x) = 1, the limit of f(x) as x approahes any real number: a. equals zero because f(x) is a constant b. equals 1 provided that the domain of f(x) is xER c. two of the other answers are correct d. does not exist

Chapter 2 Solutions

Calculus - Standalone book

Ch. 2.1 - Prob. 1PQ Ch. 2.1 - Prob. 2PQ Ch. 2.1 - Prob. 3PQ Ch. 2.1 - Prob. 4PQ Ch. 2.1 - Prob. 5PQ Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E

Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.1 - Prob. 30E Ch. 2.1 - Prob. 31E Ch. 2.1 - Prob. 32E Ch. 2.1 - Prob. 33E Ch. 2.1 - Prob. 34E Ch. 2.1 - Prob. 35E Ch. 2.1 - Prob. 36E Ch. 2.2 - Prob. 1PQ Ch. 2.2 - Prob. 2PQ Ch. 2.2 - Prob. 3PQ Ch. 2.2 - Prob. 4PQ Ch. 2.2 - Prob. 5PQ Ch. 2.2 - Prob. 6PQ Ch. 2.2 - Prob. 7PQ Ch. 2.2 - Prob. 1E Ch. 2.2 - Prob. 2E Ch. 2.2 - Prob. 3E Ch. 2.2 - Prob. 4E Ch. 2.2 - Prob. 5E Ch. 2.2 - Prob. 6E Ch. 2.2 - Prob. 7E Ch. 2.2 - Prob. 8E Ch. 2.2 - Prob. 9E Ch. 2.2 - Prob. 10E Ch. 2.2 - Prob. 11E Ch. 2.2 - Prob. 12E Ch. 2.2 - Prob. 13E Ch. 2.2 - 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