Explanation Given Information: The limit is lim x → 1 f ( x ) . The function is f ( x ) = { x 2 +1 if x ≤ 0 x if 0 < x < 1 2 x 2 − 1 if x ≥ 1 . Formula used: The lim x → c f ( x ) will exist at c if, lim x → c − f ( x ) = lim x → c + f ( x ) By the properties of limit lim x → c x can be simplified as, lim x → c x = c . Calculation: Consider the provided function, f ( x ) = { x 2 +1 if x ≤ 0 x if 0 < x < 1 2 x 2 − 1 if x ≥ 1 Consider the provided limit, lim x → 1 f ( x ) The limit from the left is represented by lim x → 1 − f ( x ) and the limit from the right is represented by lim x → 1 + f ( x ) . The lim x → c f ( x ) will exist at c = 1 when the limit from the left, that is, the values of f ( c ) but c < 1 , is equal to the limit from the right, that is, the values of f ( c ) but c > 1

WebAssign Printed Access Card for Harshbarger/Reynolds' Mathematical Applications for the Management, Life, and Social Sciences, 12th Edition, Multi-Term

12th Edition

ISBN: 9781337630542

Author: Ronald J. Harshbarger, James J. Reynolds

Publisher: Cengage Learning

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Chapter 9, Problem 27RE

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To calculate: The value of limx→1f(x) for the function, f(x)={x2+1 if x≤0x if 0<x<12x2−1 if x≥1.

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Chapter 9, Problem 26RE

Chapter 9, Problem 28RE

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Chapter 9 Solutions

WebAssign Printed Access Card for Harshbarger/Reynolds' Mathematical Applications for the Management, Life, and Social Sciences, 12th Edition, Multi-Term

Ch. 9.1 - 1. Can exist if is undefined? Ch. 9.1 - 2. Does exist if ? Ch. 9.1 - 3. Does Ch. 9.1 - 4. If , does exist? Ch. 9.1 - 5. Let (a) Does (b) Find . Ch. 9.1 - 6. Evaluate the following limits (if they...Ch. 9.1 - Prob. 7CP Ch. 9.1 - Prob. 8CP Ch. 9.1 - Prob. 9CP Ch. 9.1 - Prob. 10CP

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The value of lim x → 1 f ( x ) for the function, f ( x ) = { x 2 +1 if x ≤ 0 x if 0 < x < 1 2 x 2 − 1 if x ≥ 1 .

Solution Summary: The author explains how the value of undersetxto 1mathrmlimf(x)) can be simplified by the properties of limit.

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To calculate: The value of limx→1f(x) for the function, f(x)={x2+1 if x≤0x if 0<x<12x2−1 if x≥1.

Want to see the full answer?

Chapter 9 Solutions

The value of lim x → 1 f ( x ) for the function, f ( x ) = { x 2 +1 if x ≤ 0 x if 0 &lt; x &lt; 1 2 x 2 − 1 if x ≥ 1 .

Solution Summary: The author explains how the value of undersetxto 1mathrmlimf(x)) can be simplified by the properties of limit.

Videos

To calculate: The value of limx→1f(x) for the function, f(x)={x2+1 if x≤0x if 0<x<12x2−1 if x≥1.

Want to see the full answer?

Chapter 9 Solutions

The value of lim x → 1 f ( x ) for the function, f ( x ) = { x 2 +1 if x ≤ 0 x if 0 < x < 1 2 x 2 − 1 if x ≥ 1 .