Solutions for Finding Limits: Numerical and Graphical Approaches
Solutions to Try Its
1. a=5, f(x)=2x2−4, and L=46.
2. a. 0; b. 2; c. does not exist; d. −2; e. 0; f. does not exist; g. 4; h. 4; i. 4
3. x→0lim(4x20sin(x))=5
4. does not exist
Solutions to Odd-Numbered Exercises
1. The value of the function, the output, at x=a is f(a). When the x→alimf(x) is taken, the values of x get infinitely close to a but never equal a. As the values of x approach a from the left and right, the limit is the value that the function is approaching.
3. –4
5. –4
7. 2
9. does not exist
11. 4
13. does not exist
15. Answers will vary.
17. Answers will vary.
19. Answers will vary.
21. Answers will vary.
23. 7.38906
25. 54.59815
27. e6≈403.428794, e7≈1096.633158, en
29. x→−2limf(x)=1
31. x→3lim(x2−9x2−x−6)=65≈0.83
33. x→1lim(x2−3x+2x2−1)=−2.00
35. x→1lim(x2−3x+210−10x2)=20.00
37. x→2−1lim(4x2+4x+1x) does not exist. Function values decrease without bound as x approaches –0.5 from either left or right.
39. x→0lim3x7tanx=37
41. x→0lim4tanx2sinx=21
43. x→0limee−x21=1.0
45. x→−1−limx+1∣x+1∣=(x+1)−(x+1)=−1 and x→−1+limx+1∣x+1∣=(x+1)(x+1)=1; since the right-hand limit does not equal the left-hand limit, x→−1limx+1∣x+1∣ does not exist.
47. x→−1lim(x+1)21 does not exist. The function increases without bound as x approaches −1 from either side.
49. x→0lim1−ex25 does not exist. Function values approach 5 from the left and approach 0 from the right.
51. Through examination of the postulates and an understanding of relativistic physics, as v→c, m→∞. Take this one step further to the solution,