1. Selected Problems on Limits and Continuity

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

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

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

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

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

7. 3x = lim 2 3 sin 3x 1 = lim 2 ● 1 ● 3 = 2 ( 1 ) ( 1 ) cos 3(0) 3 2/3
● 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 = lim (x2 + a2) (x2 – a2) x2 – a2 = lim (x2 + a2) 2a2

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

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

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

13. 6. The function below is discontinuous at x = 3
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. The graph of these two functions are identical except for the hole.
This function is discontinuous at (3,5)
f (x) = 2x2 – 7x + 3
x – 3
f(x) = 2x2 – 7x if x  3
x – 3
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 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 42 – c2 c(4) + 20 42 – c2 = 4c + 20

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 – if x  5

19. lim ax2 + x – b lim 2x + b 2(2) + b a(2)2 + 2 – b 4a + 2 – b 4 + b

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 – if x  5

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

22. 4a – 20a = 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 cos x cos x ax ● lim sin x ax cos x lim sin x
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 ● ● 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 lim g(x) = 0 lim h(x) = - 8
find the following;
a) lim ( f(x) + g(x) )
x3
= lim f(x) + lim g(x)
x x3 =
= 5

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

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

31. = ( - 8 )
5 – ( - 8 ) = 13

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