Explanation Given: lim x → ∞ sin ( 1 x ) lim x → 0 sin ( 1 x ) Limits represent the tendency of a function. Limits do not exist if we cannot determine the tendency of the function to a single point. Therefore, a function does not approach a limit in the following cases, There is a jump discontinuity There is a vertical asymptote There is a violent oscillation To explain whether a given limit is present or not we need to examine if any of the aforementioned conditions apply to the given limit. Let us evaluate the given limits numerically and then graphically. To solve the limits numerically we can apply the limit chain rule. We can graph the function f ( x ) = sin ( 1 x ) to see the tendency of the function at the given values of x. The graph of the function is drawn as follows, We can see that as x gets closer to zero; the function keeps oscillating back and forth between -1 and 1. We can conclude that as x approaches 0 from the right or left, the function sin ( 1 x ) does not settle down on any value, and so the limit as x approaches 0 from the right or left does not exist. Whereas, as x gets closer to infinity on the far right side of the graph, the function sin ( 1 x ) approaches the x -axis, this means that the limit lim x → ∞ sin ( 1 x ) is 0. Formula used: The limit chain rule: If lim u → b f ( u ) = L , and lim x → a g ( x ) = b , and f ( x ) is continuous at x = b then: lim x → a f ( g ( x ) ) = L Calculations: Description lim x → ∞ sin ( 1 x ) Step 1: Apply the limit chain rule, lim u → b f ( u ) = L , and lim x → a g ( x ) = b , then: lim x → a f ( g ( x ) ) = L Here, g ( x ) = 1 x and f ( u ) = sin u lim x → ∞ g ( x )

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Fundamentals of Trigonometric Identities

Trigonometric Identities are the equalities that involve trigonometry functions and hold for all the values of variables given in the equation.

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Chapter 2.7, Problem 6PQ

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The limit limx→∞sin(1x) and explain the reason why limx→∞sin(1x) exists but limx→0sin(1x) does not exist.

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Chapter 2.7, Problem 5PQ

Chapter 2.7, Problem 1E

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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

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The limit lim x → ∞ sin ( 1 x ) and explain the reason why lim x → ∞ sin ( 1 x ) exists but lim x → 0 sin ( 1 x ) does not exist.

Solution Summary: The author evaluates the function mathrmsin(1x) and explains why it does not exist.

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The limit limx→∞sin(1x) and explain the reason why limx→∞sin(1x) exists but limx→0sin(1x) does not exist.

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Chapter 2 Solutions