MATH 219T (2223) Level N Mathematics Calculus I Course Questions Solutions (GRADED A)

Calculus I Solutions

Chapter 1 Functions, Limits, and Continuity

Section 1.1 Preliminaries

(1) 1. The function h is not defined at x  1 and x  4.
Thus, the domain of h is R  {1, 4}.
Since 2 < 1, then the rule h(x)  0 is used to evaluate h(2): h(2)  0.
Since 0 < 1, then the rule h(x)  0 is used to evaluate h(0): h(0)  0.
Since 1 < 2 < 4, then the rule h(x)  x2  1 is used to evaluate h(2):
h(2) = 22  1  3.
Since 5 > 4, then the rule h(x)  2x + 1 is used to evaluate h(5):
h(5)  2(5) + 1  11.
(2) 2. a. f (x) is many-to-one function as, for example, f (0) = 4 and f (4) = 4.
b. Let a and b be the elements in the domain of g such that g(a) = g(b), that is
1 1
 or 3a  9 = 3b  9, which simplifies to a = b. Therefore, g is a
3a  9 3b  9
one-to-one function.
(3) 3. a. f (x) = 2(x)2 + 5 + sec (x) = 2x2 + 5 + sec x = f (x) and therefore, f is even.
b. h(x) = 2(x) cos (x) = 2x cos x  h(x) and therefore, h is not even.
(4) 4. a. f (x) = (x)3 + (x) | x | = x3  x | x | = (x3 + x | x |) = f (x) and therefore, f is
odd.
b. h(x) = 7(x) + 5| x | = 7x + 5 | x |  h(x) and therefore, h is not odd.
(5) 5. Case 1: x  4  0, x  4:
x4<1x
2x < 5
x < 2.5 (rejected)
Case 2: x  4 < 0, x < 4:
4x<1x
4 < 1 (false)
Thus, the solution set of the inequality is the empty set { }.
(6) 6. Case 1: x < 3:
3x + 1 > 2(x + 3)
3x + 1 > 2x  6
x > 7
x<7
Hence, x < 3.




Level N | 1

, Calculus I



Case 2: 3  x < 1/3:
3x + 1 > 2(x + 3)
3x + 1 > 2x + 6
5x > 5
x < 1
Hence, 3  x < 1.
Case 3: x  1/3:
3x  1 > 2(x + 3)
3x  1 > 2x + 6
x>7
Hence, x > 7. Thus, the solution set of the inequality is x < 1 or x > 7.

Section 1.2 Limit of a Function


(7) 7. | 4x + 1  33 | < 0.2
| 4x  32 | < 0.2
4 | x  8 | < 0.2
| x  8 | < 0.05
Conclusion: x must be within 0.05 from 8 for h(x) to be within 0.2 from 33.

1 1 1 1 π
(8) 8. a) lim    b) lim  tan x   tan 1
x 1 x  3 1  3 2 2 x
π 4
4



 π  π
sin 2  x   sin 2  x  
(9) 9. a) limπ  2 1
 lim  2
2 2
x   π  4 x 
π
π
2
2  x  2 
2
x 
    2
π π
Let y  x  , then as x  , y  0 . This yields
2 2
2 2
1 sin 2 y 1  sin y  1  sin y  1
lim 2  lim     lim   .
4 y 0 y 4 y 0  y  4  y 0 y  4
 x  1  lim  x  1  0
2
x2  2x  1
b) lim  lim
x 1 x2 1 x 1  x  1 x  1 x 1  x  1


x3  8  x  2  x2  2 x  4  22  2  2  4 12
c) lim 2  lim   3
x 2 x  4 x 2
 x  2  x  2  2  2 4




Level N | 2

, Calculus I



 sin(sin x)   sin(sin x) 
  lim  
(10) 10. a) lim
sin(sin x)
 lim  sin x 

sin x  0
 sin x  1
 1
x 0 x x 0  x   x  1
  lim  
 sin x  x  0
 sin x 
cos x  1 (cos x  1) (cos x  1)  sin 2 x
b) lim  lim  lim
x 0 sin x x 0 sin x (cos x  1) x 0 sin x (cos x  1)

 sin x 0
lim  0
= x0 cos x  1 2

(11) 11. We must find the value of  that will ensure that 2 x  1  5  ε .
ε ε
Simplifying 2 x  1  5  ε yields 2 x  4  ε, x  2  . Therefore, taking δ 
2 2
will insure that 2 x  1  5  ε is true.

4x 8 4x 8
(12) 12. a) lim     b) lim    
x 2 x  2 0 x 2 x  2 0
4x 4x 4x
c) lim does not exist since lim  lim
x 2 x  2 x 2 x  2 x 2 x  2


4  x2 2  x 2  x
d) lim  lim  lim  2  x   2   2   4
x 2 x  2 x 2 x2 x 2

4  x2
e) lim  lim  2  x   2   2   4
x 2 x  2 x 2

4 x 2
f) lim  lim  2  x   2   2   4
x 2 x  2 x 2

x x
g) lim    0 h) lim    1
x 2  2  x 2  2 

x x x
i) As lim    lim   , lim   does not exist.
x 2  2  x 2  2   
x 2 2




Level N | 3

, Calculus I




 if n  m

 an x n  an 1 x n 1  ...  a0   an
13. [G1] lim   if n  m
(13)
x  b x m  b m 1
 m m 1 x  ...  b0   bn
 0 if n  m
3x 2  x  2
a) lim 2  3 . Answer: 3
x  x  4 x  1

2 x 3  4 x  12
b) lim is infinite. Answer: 999
x  3x 2  x
1  5 x  x5
c) lim is infinite. Answer: 999
x  2 x3
6  5x2 5
d) lim  . Answer: 5/3
x  3 x 2  7 3
x  3x  1
6 4
e) lim 7  0 . Answer: 0
x  x  7 x 3  1

24 x 2  x  12
f) lim 3  0 . Answer: 0
x  8 x  3 x  11




x 2  256  x  16  x  16   lim  x 4  
x  4  x  16 
(14) 14. [G1] lim  lim
x 16 x  4 x16 x 4 x 16
x 4
 lim
x 16
 
x  4  x  16    8 32   256
(15) 15. [G1]
2  2 cos x 2 1  cos x   1  cos x  2sin 2 x
a) lim  lim    lim
x 0 sin 2 x x 0 sin 2 x  1  cos x  x 0 2sin x cos x (1  cos x)
sin x 0
 lim   0 . Answer: 0
x 0 cos x (1  cos x) 2
sin x
b) lim  0 . Answer: 0
x 0 cos x 1  cos x 

cos x 1
c) lim  . Answer: ½
x 0 1  cos x 2
2
d) lim  1 . Answer: 1
x 0 1  cos x

5sin x  sin10 x 5 sin x 1  sin10 x  5 10
e) lim  lim  lim10    1  (1)  5
x 0 3x 3 x0 x 3 x0  10 x  3 3
Answer: 5




Level N | 4