Calculus
7th Edition
ISBN: 9781524916817
Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE
Publisher: Kendall Hunt Publishing
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Question
Chapter 4, Problem 86SP
To determine
Find out the solution of the given limit problem.
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Check out a sample textbook solutionStudents have asked these similar questions
lim x → 9− f(x) = 2 and lim x → 9+ f(x) = 4.
As x approaches 9 from the right, f(x) approaches 2. As x approaches 9 from the left, f(x) approaches 4.
As x approaches 9 from the left, f(x) approaches 2. As x approaches 9 from the right, f(x) approaches 4.
As x approaches 9, f(x) approaches 4, but f(9) = 2.
As x approaches 9, f(x) approaches 2, but f(9) = 4.
In this situation is it possible that
lim x → 9 f(x)
exists? Explain.
Yes, f(x) could have a hole at (9, 2) and be defined such that f(9) = 4.
Yes, f(x) could have a hole at (9, 4) and be defined such that f(9) = 2.
Yes, if f(x) has a vertical asymptote at x = 9, it can be defined such that lim x→9− f(x) = 2, lim x→9+ f(x) = 4, and lim x→9 f(x) exists.
No, lim x→9 f(x) cannot exist if lim x→9− f(x) ≠ lim x→9+ f(x).
(a) What is wrong with the following equation?
x2 + x − 12
x − 3
= x + 4
(x − 3)(x + 4) ≠ x2 + x − 12
The left-hand side is not defined for x = 0, but the right-hand side is. The left-hand side is not defined for x = 3, but the right-hand side is.None of these — the equation is correct.
(b) In view of part (a), explain why the equation
lim x → 3
x2 + x − 12
x − 3
= lim x → 3 (x + 4)
is correct.
Since
x2 + x − 12
x − 3
and x + 4 are both continuous, the equation follows.
Since the equation holds for all x ≠ 3, it follows that both sides of the equation approach the same limit as x → 3. This equation follows from the fact that the equation in part (a) is correct.None of these — the equation is not correct.
lim x->0 ((x+k)2 -x2/k)
Chapter 4 Solutions
Calculus
Ch. 4.1 - Prob. 1PSCh. 4.1 - Prob. 2PSCh. 4.1 - Prob. 3PSCh. 4.1 - Prob. 4PSCh. 4.1 - Prob. 5PSCh. 4.1 - Prob. 6PSCh. 4.1 - Prob. 7PSCh. 4.1 - Prob. 8PSCh. 4.1 - Prob. 9PSCh. 4.1 - Prob. 10PS
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Prob. 58PSCh. 4.7 - Prob. 59PSCh. 4.7 - Prob. 60PSCh. 4 - Prob. 1PECh. 4 - Prob. 2PECh. 4 - Prob. 3PECh. 4 - Prob. 4PECh. 4 - Prob. 5PECh. 4 - Prob. 6PECh. 4 - Prob. 7PECh. 4 - Prob. 8PECh. 4 - Prob. 9PECh. 4 - Prob. 10PECh. 4 - Prob. 11PECh. 4 - Prob. 12PECh. 4 - Prob. 13PECh. 4 - Prob. 14PECh. 4 - Prob. 15PECh. 4 - Prob. 16PECh. 4 - Prob. 17PECh. 4 - Prob. 18PECh. 4 - Prob. 19PECh. 4 - Prob. 20PECh. 4 - Prob. 21PECh. 4 - Prob. 22PECh. 4 - Prob. 23PECh. 4 - Prob. 24PECh. 4 - Prob. 25PECh. 4 - Prob. 26PECh. 4 - Prob. 27PECh. 4 - Prob. 28PECh. 4 - Prob. 29PECh. 4 - Prob. 30PECh. 4 - Prob. 1SPCh. 4 - Prob. 2SPCh. 4 - Prob. 3SPCh. 4 - Prob. 4SPCh. 4 - Prob. 5SPCh. 4 - Prob. 6SPCh. 4 - Prob. 7SPCh. 4 - Prob. 8SPCh. 4 - Prob. 9SPCh. 4 - Prob. 10SPCh. 4 - Prob. 11SPCh. 4 - Prob. 12SPCh. 4 - Prob. 13SPCh. 4 - Prob. 14SPCh. 4 - Prob. 15SPCh. 4 - Prob. 16SPCh. 4 - Prob. 17SPCh. 4 - Prob. 18SPCh. 4 - Prob. 19SPCh. 4 - Prob. 20SPCh. 4 - Prob. 21SPCh. 4 - Prob. 22SPCh. 4 - Prob. 23SPCh. 4 - Prob. 24SPCh. 4 - 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