Avon High School
AP Calculus AB
Name __________________________________
Period _____
Score ______ / 10
Skill Builder: Topic 1.6 – Determining Limits Using Algebraic Manipulation
Directions: Beginning in the first cell, find the answer. Search for that answer elsewhere in the
document, mark that cell #2 and work the problem in that box. Process in this manner until you
complete the circuit. Show all pertinent work. Calculators may not be used.
Note: This assignment contains two separate circuits.
Circuit 1
#1
−3
Answer:
8
# _5_
Answer: 0
2x − 2
x →1 x − 1
lim12
lim
x→5
12 (The limit of a constant is a constant.)
lim
x →1
# _8_
Answer:
3
8
 x − 1, x  3
If f ( x) = 
, find lim f ( x).
x →3
2 x − 3, x  3
lim ( x − 1) = 3 − 1 = 2
x →3−
lim (2(3) − 3) = 6 − 3 = 3
x →3+
lim f ( x) does not exist
x →3
# _3_
Answer: 8
2x − 2
2( x − 1)
= lim
= lim 2 = 2
x
→
1
x →1
x −1
x −1
# _6_
s 2 + 6s + 5
s →−1 s 2 − 3s − 4
s 2 + 6s + 5
( s + 5)( s + 1)
lim 2
= lim
s →−1 s − 3s − 4
s →−1 ( s − 4)( s + 1)
s+5
= lim
s →−1 s − 4
4
=−
5
lim
# _11_
1 1
+
lim 2 t
t→−2 2 + t
lim (3 x 2 − 4 x − 1)
x →5
lim ( 3x 2 − 4 x − 1) = 3(5) 2 − 4(5) − 1
x →5
= 75 − 20 − 1
= 54
Answer: 2
t
2
+
= lim 2t 2t
x→ −2 2 + t
t+2 1
= lim

x → − 2 2t
2+t
1
1
= lim = −
x → − 2 2t
4
Answer:
1
4
Answer: −1
# _10_
x+4 −2
x
lim
x →0
x →0
= lim
x →0
x
= lim
x →0
(
x+4+2
lim 4 x = 4(2) = 8
x →2
)
1
1
=
x+4+2 4
Answer:
54
x 2 + 4 x + 4 (−2) 2 + 4(−2) + 4
=
x →− 2
x2
(−2) 2
4−8+ 4
=
4
0
= =0
4
x3 + x 2 − 4 x − 4
x 4 − 16
x 2 ( x + 1) − 4( x + 1)
x →2
x 4 − 16
( x + 1)( x 2 − 4)
= lim 2
x → 2 ( x − 4)( x 2 + 4)
x +1
= lim 2
x →2 x + 4
3
=
8
= lim
Answer:
DNE
lim (cos  − sin  ) = −1
lim
# _7_
# _9_
cos x − sin  , x  
If f ( x) = 
, find lim f ( x).
x →
 x −  − 1, x  
x2 + 4 x + 4
x →− 2
x2
lim
x→ 2
12
x →2
# _4__
lim
Answer:
lim 4 x
x+4 −2 x+4 +2

x
x+4 +2
x+4−4
= lim
# _2__
Answer:
x → −
lim ( −  − 1) = −1
x → +
lim f ( x) = −1
x →
−4
5
# _12_
5 x 2 − 2 x3 + 2 x − 5
lim
x →1 4 x 2 + 6 x − 10 x 3
−2 x 3 + 5 x 2 + 2 x − 5
lim
x →1 −10 x 3 + 4 x 2 + 6 x
− x 2 (2 x − 5) + 1(2 x − 5)
= lim
x →1
−2 x(5 x 2 − 2 x − 3)
(2 x − 5)(1 − x 2 )
= lim
x →1 −2 x (5 x + 3)( x − 1)
−(2 x − 5)( x − 1)( x + 1)
= lim
x →1
−2 x(5 x + 3)( x − 1)
(2 x − 5)( x + 1) (−3)(2)
3
= lim
=
=−
x →1
2 x(5 x + 3)
2(8)
8
Answer:
−1
4
Circuit 2
#1
Answer:
−1
lim 
# _5_
lim
x →0
x→4
=
Answer:
4
3
x 2 − 16
x−4
( x − 4)( x + 4)
x−4
= lim( x + 4)
= lim
x→4
x→4
=8
# _7_
Answer:
12
x2 − 4x + 4
lim 2
x→ 2 x + x − 6
lim
x→4
lim ( 5 x3 − 7 x 2 + 2 x − 2 )
x →1
= 5(1)3 − 7(1) 2 + 21 − 2
= −2
Answer: 11
2x − 4
x −1
2(4) − 4
4 −1
4
=
3
( x − 2)( x − 2)
x → 2 ( x + 3)( x − 2)
x−2
= lim
x→2 x + 3
0
= =0
5
=
= lim
# _2_
# _4_
Answer:
𝜋
# _12_
Answer: 2 3
1
 1

lim 
+ 2

x →1 x − 1
x − 3x + 2 

x 2 − 3x + 2
x −1
+
2
x →1 ( x − 1)( x − 3 x + 2)
( x − 1)( x 2 − 3 x + 2)
= lim
x2 − 2x + 1
x →1 ( x − 1)( x 2 − 3 x + 2)
( x − 1)( x − 1)
= lim
x →1 ( x − 1)( x − 1)( x − 2)
1
= lim
x →1 x − 2
= −1
= lim
# _11_
lim
h →0
1
4
# _10_
h2
h2 + h + 3 − h + 3
h2
= lim
h2 + h + 3 − h + 3
h →0
= lim
Answer:
h2
h →0
= lim
h →0
(
lim

h2 + h + 3 + h + 3
h2 + h + 3 + h + 3
h2 + h + 3 + h + 3
)
x →0
x2 + x + 4 − 2
x2 + x + 4 + 2

x2 + x
( x 2 + x) x 2 + x + 4 + 2
= lim
(
2
= lim
h2 + h + 3 + h + 3
1
x →0
Answer: 8
= lim
Answer: 0
= 2(32 + 3(3) + 9)
= 54
Answer: 54
 x 2 + x, x  −1
If f ( x) =  − x
, find lim f ( x).
x →−1
−2 , x  −1
lim f ( x) does not exist
x2 + x + 4 + 2
x →3
= (−2) 2 − 2(−2) + 4
= 4 + 4 + 4 = 12
lim (−2− ( −1) ) = −2
1
2( x − 3)( x 2 + 3 x + 9)
x →3
x −3
= lim  2( x 2 + 3 x + 9) 
x →− 2
x →−1+
)
2 x3 − 54
x →3
x −3
(t + 2)(t 2 − 2t + 4)
= lim
x →− 2
t+2
2
= lim (t − 2t + 4)
lim ( (−1) 2 + (−1) ) = 0
x2 + x + 4 + 2
# _8_
lim
x →−1−
(
)
1
4
t3 + 8
lim
t →− 2 t + 2
# _9_
( x 2 + x)
x →0
=
# _6_
x2 + x + 4 − 4
= lim
=2 3
x →−1
x2 + x + 4 − 2
x2 + x
x →0
h + h + 3 − (h + 3)
Answer: DNE
# _3__
lim (3 y 4 − 6 y 3 − 2 y )
y →−1
= 3(−1) 4 − 6(−1)3 − 2(−1)
= 3+ 6+ 2
= 11
Answer: −2