Selected Problems on Limits and Continuity

1. Selected Problems on Limits and Continuity

2. 1. lim 2x3 – 5x2 + 3x x1 3x2 – 5x + 2 = lim x(2x2 – 5x + 3) x  1 3x2 – 5x + 2 = lim x (2x – 3)(x – 1) x 1 (3x – 2)(x – 1) = (1) ( 2 (1) – 3) = (1)(2 – 3 ) - 1 (3 – 2 ) ( 3(1) – 2 )

3. x2 + 3 + 2 2. lim x – 1 ● x1 x2 + 3 - 2 x2 + 3 + 2 = lim (x – 1) ( x2 + 3 + 2) x1 x2 + 3 − 4 = lim (x – 1) ( x2 + 3 + 2 ) x1 x2 – 1

4. = lim (x – 1) ( x2 + 3 + 2 ) x1 (x + 1)(x – 1 ) = lim ( x2 + 3 + 2 ) x1 x + 1 = 4 + 2 = 12 + 3 + 2 2 1 + 1 = 2

5. 3. lim 2x x 0 tan 3x = lim 2x x  0 sin 3x cos 3x cos 3x = lim 2x ● sin 3x x 0 1

6. = lim 2x cos 3x x 0 sin 3x = lim 2 x ● cos 3x ● x 0 sin 3x x = lim 2 ● 3 ● cos 3x ● 3 x 0 sin 3x

7. 3x = lim 2 ● 1 ● cos 3x ● 3 x 0 sin 3x 1 ● cos 3x = lim 2 ● 1 ● 3 x 0 = 2 ( 1 ) ( 1 ) cos 3(0) 3 2/3 = 2 ● cos 0 = 2 ● 1 3 3

8. 4. lim x4 – a4 xa x2 – a2 = lim (x2 + a2) (x2 – a2) x  a x2 – a2 = lim (x2 + a2) x  a 2a2 = a2 + a2

9. 5. Find the value of a and b such that lim a + bx - 3 3 = x  0 x

10. a + bx + 3 lim a + bx - 3 ● x  0 x a + bx + 3 3 lim a + bx - 3 = 3 x  0 (x) ( a + bx + 3 ) lim 3 + bx - 3 = 3 x  0 (x) ( 3 + bx + 3 )

11. lim bx = 3 x  0 (x) ( 3 + bx + 3 ) lim b = 3 x  0 ( 3 + bx + 3 ) b = 3 ( 3 + b(0) + 3 )

12. b = 3 3 + 3 b = 3 2 3 1 b = 2 9 a = 3 b = 6

13. 6. The function below is discontinuous at x = 3. Redefine this function to make it continuous. f (x) = 2x2 – 7x + 3 x – 3 Note: To make a function continuous, we are essentially going to fill up the hole.

14. f (x) = 2x2 – 7x + 3 x – 3 f(x) = (2x – 1 )(x – 3 ) x – 3 f(x) = 2x – 1 f(3) = 2(3) – 1 x – 3 = 0 f(3) = 5 x = 3 hole: (3, 5)

15. This function is discontinuous at (3,5) f (x) = 2x2 – 7x + 3 x – 3 f(x) = 2x2 – 7x + 3 if x  3 x – 3 5 if x = 3 The graph of these two functions are identical except for the hole.

16. 7. The function below is continuous. Find the value of the constant c. f(x) = x2 – c2 if x < 4 cx + 20 if x  4 Since each part of these function is a polynomial function, then each of them is continuous. Thus the only possible point of discontinuity is at the boundary point.

17. lim x2 – c2 lim cx + 20 x4 x4 42 – c2 c(4) + 20 42 – c2 = 4c + 20 16 – c2 = 4c + 20 16 – c2 – 4c – 20 = 0 - c2 – 4c – 4 = 0 c2 + 4c + 4 = 0 (c + 2)(c + 2) = 0 c = - 2

18. 8. Find the values of a and b such if the function f below is continuous. f(x) = ax2 + x – b if x  2 2x + b if 2 < x < 5 2ax – 7 if x  5

19. lim ax2 + x – b lim 2x + b x2 x2 2(2) + b a(2)2 + 2 – b 4a + 2 – b 4 + b 4a + 2 – b = 4 + b 4a + 2 – b – 4 – b = 0 4a – 2b - 2 = 0

20. 8. Find the values of a and b such if the function f below is continuous. f(x) = ax2 + x – b if x  2 2x + b if 2 < x < 5 2ax – 7 if x  5

21. lim 2x + b lim 2ax - 7 x5 x5 2(5) + b 2a(5) – 7 10 + b = 10a – 7 b = 10a – 7 – 10 b = 10a - 17 2a – 2b - 2 = 0 4a – 2 (10a – 17 ) - 2 = 0

22. 4a – 20a + 34 - 2 = 0 - 16a + 32 = 0 - 16a = - 32 a = - 32 - 16 a = 2 b = 10a - 17 b = 20 – 17 b = 3 b = 10(2) – 17

23. 9. Find the value of a such that the following is continuous. f(x) = ax if x  0 tan x x2 – 2 if x < 0

24. lim x2 - 2 lim ax = x0 tan x x0 ? = - 2

25. ax lim ax lim = sin x x0 x 0 tan x cos x cos x ax ● lim sin x x0 ax cos x lim x0 sin x a ● x ● cos x lim x0 sin x

26. a ● 1 ● cos x lim x0 = a ( 1 ) ( cos 0) = a ( 1 ) ( 1 ) = a

27. lim ax lim a2 - 2 = x0 tan x x0 ? = - 2 a = - 2

28. 10. Given that lim f(x) = 5 x3 lim g(x) = 0 lim h(x) = - 8 x3 x3 find the following; a) lim ( f(x) + g(x) ) x3 = lim f(x) + lim g(x) x3 x3 = 5 + 0 = 5

29. 10. Given that lim f(x) = 5 x3 lim g(x) = 0 lim h(x) = - 8 x3 x3 find the following; b) lim ( f(x) / h(x) ) x3 = lim f(x) / lim h(x) x3 x3 = 5/-8

30. 10. Given that lim f(x) = 5 x3 lim g(x) = 0 lim h(x) = - 8 x3 x3 find the following; c) lim 2 h(x) x3 f(x) – h(x) = 2 lim h(x) x3 lim f(x) - lim h(x) x3 x3

31. = 2 ( - 8 ) 5 – ( - 8 ) = - 16 13

32. 10. Given that lim f(x) = 5 x3 lim g(x) = 0 lim h(x) = - 8 x3 x3 find the following; d) lim x2 f(x) x3 = lim x2 ● lim f (x ) x3 x3 = 32 ● 5 = 45