Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)
10th Edition
ISBN: 9781305657861
Author: Soo T. Tan
Publisher: Cengage Learning
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Chapter B, Problem 7E
To determine
To evaluate: The value of
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Check out a sample textbook solutionStudents have asked these similar questions
In Exercises 7–10, use the formal definition of limit to prove that the function is continuous at c.
Find the limits in Exercises 33–35 Are the functions continuous at the point being approached?
Divide numerator and denominator by the highest power of x in the
denominator and proceed from there. Find the limits in Exercises 23–36.
8x² – 3
1/3
x² +
8x2 – 3
23. lim
x→0 V 2x? + x
24. lim
x--00
\5
x² – 5x
25. lim
x→-ox2 + 7x
26. lim
x→ o Vx + x – 2
2Vx + x!
2 + Vx
27. lim
28. lim
x00 2 - Vx
Зх — 7
Vx – Vĩ
VI + Vĩ
x' + x4
29. lim
30. lim
x→-00
VI
lim
32.
x--0 2x + x²/3 – 4
2x/3 – x'/3 + 7
5x + 3
31. lim
x-00 8/5 + 3x + Vx
Vx² + 1
Vx² + 1
33. lim
x00 x + 1
34. lim
x→-0 x + 1
4 - 3x3
lim
35. lim
x→∞ V4x? + 25
36.
x→-0 Vx6 + 9
Chapter B Solutions
Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)
Ch. B - Prob. 1ECh. B - Prob. 2ECh. B - Prob. 3ECh. B - Prob. 4ECh. B - Prob. 5ECh. B - Prob. 6ECh. B - Prob. 7ECh. B - Prob. 8ECh. B - Prob. 9ECh. B - Prob. 10E
Ch. B - Prob. 11ECh. B - Prob. 12ECh. B - Prob. 13ECh. B - Prob. 14ECh. B - Prob. 15ECh. B - Prob. 16ECh. B - Prob. 17ECh. B - Prob. 18ECh. B - Prob. 19ECh. B - Prob. 20ECh. B - Prob. 21ECh. B - Prob. 22ECh. B - Prob. 23ECh. B - Prob. 24ECh. B - Prob. 25ECh. B - Prob. 26ECh. B - Prob. 27ECh. B - Prob. 28ECh. B - Prob. 29ECh. B - Prob. 30ECh. B - Prob. 31ECh. B - Prob. 32ECh. B - Prob. 33ECh. B - Prob. 34E
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- In Exercises 11–50, evaluate the limit if it exists. If not, determine whether the one-sided limits exist. For limits that don't exist indicate whether they can be expressed as = -o or = ∞.arrow_forwardFind the limits in Exercises 13–20. (If in doubt, look at the function's graph.) 14. lim cos!x 13. lim sinx x→I- 15. lim tan-x 17. lim sec-lx 16. lim tan 18. lim sec-lx 19. lim csc¬lx 20. lim csc-lxarrow_forwardIn Exercises 49–54, show that the limits do not exist.arrow_forward
- Find the limits in Exercises 25–30.arrow_forwardIn Exercises 3–8, find the limit of each function (a) as x→∞ and (b) as x→ -. (You may wish to visualize your answer with a graphing calculator or computer.) 2 4. f(x) = 7 x² 3. f(x) = 3 5. g(x) 6. g(x) 8 – (5/x²) 2 + (1/x) -5 + (7/x) 3 - (1/x²) 3 - (2/x) 7. h(x) 8. h(x) 4 + (V2/x²)arrow_forwardProve the limit statements in Exercises 37–50.arrow_forward
- Evaluate the limits in Exercises 55–60. Use the theorems in this section to justify each step of your work. 55. lim k→o 2k sin k (Hint: Use the Squeeze Theorem.) (1+:)(--}) 56. lim k00 k – 7 k3 +1 57. lim k00 3k +5 58. lim k2 59. lim (Vk2 + k – k) 60. lim (Vk3 +k+1 - vk3 – k – 1) (Hint: Write as a quo- k00 tient and rationalize the numerator.)arrow_forwardFind the limits in Exercises 125–132. sin x 3x – tan 7x 125. lim x0 2x? – x 126. lim 2x x→0 sin (sin 0) sin r 127. lim 128. lim 0 tan 2r 4 tan? 0 + tan 0 lim + 1 129. tan² 0 + 5 0→(T/2) – 2 cot? 0 1 130. lim 60* 5 cot? 0 – 7 cot 0 – 8 1 - cos 0 02 x sin x 131. lim x→0 2 – 2 cos x 132. lim 0→0arrow_forwardEach of Exercises 31–36 gives a function ƒ(x), a point c, and a positivenumber P. Find L = lim xSc ƒ(x). Then find a number d 7 0 suchthat for all x 0 < x - c < d => ƒ(x) - L < P.arrow_forward
- Each of Exercises 31–36 gives a function ƒ(x), a point c, and a positivenumber P. Find L = lim xSc ƒ(x). Then find a number d 7 0 suchthat for all x0 < x - c < d => ƒ(x) - L < P.arrow_forwardEach of Exercises 31–36 gives a function ƒ(x), a point c, and a positivenumber P. Find L = lim xSc ƒ(x). Then find a number d 7 0 suchthat for all x0 < x - c < d => ƒ(x) - L < P. ƒ(x) = -3x - 2, c = -1, P = 0.03arrow_forwardIn Exercises 11–20, verify each limit using the limit definition. For ex- ample, in Exercise 11, show that |2x – 6| can be made as small as desired by taking x close to 3. - 11. lim 2x x→3 12. lim 4 = 4 x→3 6 13. lim (4x + 3) = 11 x→2 14. lim (5x – 7) = 8 x→3 %3D 15. lim (-2x) = -18 x→9 16. lim (1 — 2х) — 11 x→-5 17. lim x? = 0 18. lim (2x2 +4) = 4 x→0 x→0 19. lim (x2 + 2x + 3) = 3 x→0 20. lim (x + 9) = 9 x→0arrow_forward
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