Explanation Given information: The figure is, Figure (1) Calculation: Since, cos 0 = 1 so, | − 1 | ≤ cos x ≤ | 1 | for 0 ≤ | x | < π 2 Thus, lim x → 0 | − 1 | ≤ lim x → 0 cos x ≤ lim x → 0 | 1 | Resulting, 1 ≤ lim x → 0 cos x ≤ 1 lim x → 0 cos x = 1 Hence, the limit exists and, lim x → 0 cos x = cos 0 = 1 Then, f ( x ) = cos x is continuous at x = 0 . Let x 0 be an arbitrary real number

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topic…

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Chapter 2, Problem 91SP

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To show: The proof that cosx is continuous for all real x .

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Chapter 2, Problem 90SP

Chapter 2, Problem 92SP

Chapter 2 Solutions

Calculus

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