AP Calculus
L. 2.1 #1-4, 8-28E, 39-43
In Exercises 1-4, an object dropped from rest from the top of a tall building falls
y = 16t2 feet in the first t seconds.
1) Find the average speed during the first 3 seconds of fall.
2) Find the average speed during the first 4 seconds of fall.
3) Find the speed of the object at t = 3 seconds and confirm your answer
algebraically.
4) Find the speed of the object at t = 4 seconds and confirm your answer
algebraically.
Determine the limit by substitution. Support graphically.
8) lim (x + 3)1998
x¬-4
y2 + 5y + 6
y+2
10)
lim
y¬2
12)
lim int x (aka lim [x])
y¬1/2
y¬1/2
[x] is the greatest integer less than or equal to x
14)
lim x + 3
x¬2
1
In Exercise 16 and 18, explain why you cannot use substitution to determine
the limit. Find the limit if it exists.
16)
18)
1
lim
x¬0 x2
lim
x¬0
(4 + x)2 - 16
x
Determine the limit graphically. Confirm algebraically.
20)
lim
t¬2
t2 - 3t + 2
t2 - 4
1
1
2+x 2
22)
lim
x¬0
24)
lim
x¬0
sin 2x
(Difficult...)
x
26)
lim
x¬0
x + sin x
(Similar to #24)
x
28)
x
3 sin 4x
lim
x¬0 sin 3x
2
In Exercises 39-44, use the graph to estimate the limits and value of the
function, or explain why the limits do not exist.
39)
a) lim f(x)
x¬3b)
lim f(x)
x¬3+
c)
lim f(x)
x¬3
d) f(3)
40)
a) lim g(t)
t¬-4b)
lim g(t)
t¬-4+
c)
lim g(t)
t¬-4
d) g(-4)
41)
a) lim f(h)
h¬0b)
lim f(h)
h¬0+
c)
lim f(h)
h¬0
d) f(0)
3
42)
a) lim p(s)
s¬-2b)
lim p(s)
s¬-2+
c)
lim p(s)
s¬-2
d) p(-2)
43)
a) lim F(x)
x¬0b)
lim F(x)
x¬0+
c)
lim F(x)
x¬0
d) F(-2)
4
Answer Key
Testname: L. 2.1 SOLUTIONS
1)
∆y 16(3)2 - 16(0)2
=
3-0
∆t
∆y 144
=
= 48 ft/sec
3
∆t
OR
Total distance = y = 16(3)2 = 144 feet
Total time = 3 seconds
144
= 48 ft/sec
Average speed =
3
2)
∆y 16(4)2 - 16(0)2
=
4-0
∆t
∆y 256
=
= 48 ft/sec
4
∆t
OR
Total distance = y = 16(4)2 = 256 feet
Total time = 4 seconds
256
= 64 ft/sec
Average speed =
4
3)
∆y 16(3 + h)2 - 16(3)2
=
h
∆t
∆y 16(9 + 6h + h2) - 144
=
h
∆t
∆y 144 + 96h + 16h2 - 144
=
= 96 + 16h
h
∆t
As h approaches 0,
∆y
= 96 ft/sec
∆t
5
Answer Key
Testname: L. 2.1 SOLUTIONS
4)
∆y 16(4 + h)2 - 16(4)2
=
h
∆t
∆y 16(16 + 8h + h2) - 256
=
h
∆t
∆y 256 + 128h + 16h2 - 256
=
= 128 + 16h
h
∆t
As h approaches 0,
8)
10)
∆y
= 128 ft/sec
∆t
lim (x + 3)1998 = (-4 + 3)1998 = (-1)1998 = 1
x¬-4
lim
y¬2
y2 + 5y + 6 (2)2 + 5(2) + 6 4 + 10 + 6
=
=
=5
y+2
2+2
4
10
y
8
6
4
(2, 5)
2
-10 -8 -6 -4 -2
-2
2
4
6
8
10 x
-4
-6
-8
-10
6
Answer Key
Testname: L. 2.1 SOLUTIONS
12)
lim [x] = 0
y¬1/2
y
x
(0.5, 0)
14)
lim x + 3 = 2 + 3 = 5
x¬2
4
y
3
2
1
-4 -3 -2 -1
-1
1
2
3
4
x
-2
-3
-4
16)
1
1
lim
= is undefined. There is no limit...yet.
x¬0 x2 0
7
Answer Key
Testname: L. 2.1 SOLUTIONS
18)
lim
x¬0
(4 + x)2 - 16 (4 + 0)2 - 16 0
=
= appears undefined
0
0
x
however...by simplifying....
(4 + x)2 - 16
16 + 8x + x2 - 16
lim
= lim
x
x
x¬0
x¬0
= lim
x¬0
20)
8x + x2
= lim 8 + x = 8
x
x¬0
lim
t¬2
t2 - 3t + 2 1
=
4
t2 - 4
Graphically:
y
2
1
-2
-1
1
-1
2
x
(0, -.5)
-2
Algebraically:
t2 - 3t + 2
(t - 2)(t - 1)
t-1 1
lim
= lim
= lim
=
2
4
(t
2)(t
+
2)
t
+
2
t
4
t¬2
t¬2
t¬2
8
Answer Key
Testname: L. 2.1 SOLUTIONS
22)
1
1
2+x 2
lim
x¬0
x
=-
1
4
Graphically:
y
2
1
-2
-1
1
(0, -0.25)
2
x
-1
-2
Algebraically:
1
2 - (2 + x)
-x
1
2(2 + x)
4 + 2x
2+x 2
= lim
= lim
lim
x
x
x
x¬0
x¬0
x¬0
lim
x¬0
-x
4 + 2x
x
-x
-1
1
= lim
= lim
=4
x¬0 (4 + 2x)x x¬0 (4 + 2x)
9
Answer Key
Testname: L. 2.1 SOLUTIONS
24)
lim
x¬0
sin 2x
=2
x
Graphically:
3
2
y
(0, 2)
1
-3∏
4
-∏
2
-∏
4
-1
∏
4
∏
2
x
-2
-3
Algebraically:
sin 2x
2 sinx cosx
sin x
lim
= lim
= lim
œ lim 2cos x
x
x
x
x¬0
x¬0
x¬0
x¬0
= 1 œ 2 cos 0 = 1 œ 2(1) = 2
10
Answer Key
Testname: L. 2.1 SOLUTIONS
26)
lim
x¬0
x + sin x
=2
x
Graphically:
3
2
y
(0, 2)
1
-3∏
4
-∏
2
-∏
4
-1
∏
4
∏
2
x
-2
-3
Algebraically:
x + sin x
x
sin x
lim
= lim
+ lim
x
x¬0
x¬0 x x¬0 x
= 1+ 1 = 2
11
Answer Key
Testname: L. 2.1 SOLUTIONS
3 sin 4x
=4
28) lim
sin
3x
x¬0
Graphically:
5
4
y
(0, 4)
3
2
1
-3∏
4
-∏
2
-∏
4 -1
-2
∏
4
∏
2
3∏
4
x
-3
-4
-5
Algebraically:
Save this for another day...
39)
a) lim f(x) = 3
x¬3b)
lim f(x) = -2
x¬3+
c)
lim f(x) No Limit. L/R hand limits are different
x¬3
d) f(3) = 1
12
Answer Key
Testname: L. 2.1 SOLUTIONS
40)
a) lim g(t) = 5
t¬-4b)
lim g(t) = 2
t¬-4+
c)
lim g(t) No Limit. L/R hand limits are different
t¬-4
d) g(-4) = 2
41)
a) lim f(h) = -4
h¬0b)
lim f(h) = -4
h¬0+
c)
lim f(h) = -4
h¬0
d) f(0) = -4
42)
a) lim p(s) = 3
s¬-2b)
lim p(s) = 3
s¬-2+
c)
lim p(s) = 3
s¬-2
d) p(-2) = 3
13
Answer Key
Testname: L. 2.1 SOLUTIONS
43)
a) lim F(x) = 4
x¬0b)
lim F(x) = -3
x¬0+
c)
lim F(x) No Limit. L/R hand limits are different
x¬0
d) F(-2) = 4
14