Lesson 20: Derivatives and the Shapes of Curves (handout)
Lesson 4: Calculating Limits (Section 21 slides)
1. Section 1.4
Calculating Limits
V63.0121.021, Calculus I
New York University
September 16, 2010
Announcements
First written homework due today (put it in the envelope) Remember
to put your lecture and recitation section numbers on your paper
2. Announcements
First written homework due
today (put it in the
envelope) Remember to put
your lecture and recitation
section numbers on your
paper
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 2 / 45
3. Yoda on teaching a concepts course
“You must unlearn what you have learned.”
In other words, we are building up concepts and allowing ourselves only to
speak in terms of what we personally have produced.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 3 / 45
4. Objectives
Know basic limits like
lim x = a and lim c = c.
x→a x→a
Use the limit laws to
compute elementary limits.
Use algebra to simplify
limits.
Understand and state the
Squeeze Theorem.
Use the Squeeze Theorem to
demonstrate a limit.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 4 / 45
5. Outline
Recall: The concept of limit
Basic Limits
Limit Laws
The direct substitution property
Limits with Algebra
Two more limit theorems
Two important trigonometric limits
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 5 / 45
6. Heuristic Definition of a Limit
Definition
We write
lim f (x) = L
x→a
and say
“the limit of f (x), as x approaches a, equals L”
if we can make the values of f (x) arbitrarily close to L (as close to L as we
like) by taking x to be sufficiently close to a (on either side of a) but not
equal to a.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 6 / 45
7. The error-tolerance game
A game between two players (Dana and Emerson) to decide if a limit
lim f (x) exists.
x→a
Step 1 Dana proposes L to be the limit.
Step 2 Emerson challenges with an “error” level around L.
Step 3 Dana chooses a “tolerance” level around a so that points x within
that tolerance of a (not counting a itself) are taken to values y
within the error level of L. If Dana cannot, Emerson wins and the
limit cannot be L.
Step 4 If Dana’s move is a good one, Emerson can challenge again or give
up. If Emerson gives up, Dana wins and the limit is L.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 7 / 45
8. The error-tolerance game
L
a
To be legit, the part of the graph inside the blue (vertical) strip must
also be inside the green (horizontal) strip.
If Emerson shrinks the error, Dana can still win.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 8 / 45
9. Limit FAIL: Jump
y
1
x
Part of graph in-
−1 side blue is not
inside green
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 9 / 45
10. Limit FAIL: Jump
y
Part of graph in-
side blue is not 1
inside green
x
−1
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 9 / 45
11. Limit FAIL: Jump
y
Part of graph in-
side blue is not 1
inside green
x
−1
|x|
So lim does not exist.
x→0 x
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 9 / 45
12. Limit FAIL: unboundedness
y
1
lim+ does not exist
x→0 x
because the function is
unbounded near 0
L?
x
0
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 10 / 45
13. Limit EPIC FAIL
π
Here is a graph of the function f (x) = sin :
x
y
1
x
−1
For every y in [−1, 1], there are infinitely many points x arbitrarily close to
zero where f (x) = y . So lim f (x) cannot exist.
x→0
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 11 / 45
14. Outline
Recall: The concept of limit
Basic Limits
Limit Laws
The direct substitution property
Limits with Algebra
Two more limit theorems
Two important trigonometric limits
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 12 / 45
15. Really basic limits
Fact
Let c be a constant and a a real number.
(i) lim x = a
x→a
(ii) lim c = c
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 13 / 45
16. Really basic limits
Fact
Let c be a constant and a a real number.
(i) lim x = a
x→a
(ii) lim c = c
x→a
Proof.
The first is tautological, the second is trivial.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 13 / 45
22. ET game for f (x) = x
y
a
x
a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 14 / 45
23. ET game for f (x) = x
y
a
x
a
Setting error equal to tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 14 / 45
30. ET game for f (x) = c
y
c
x
a
any tolerance works!
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 15 / 45
31. Really basic limits
Fact
Let c be a constant and a a real number.
(i) lim x = a
x→a
(ii) lim c = c
x→a
Proof.
The first is tautological, the second is trivial.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 16 / 45
32. Outline
Recall: The concept of limit
Basic Limits
Limit Laws
The direct substitution property
Limits with Algebra
Two more limit theorems
Two important trigonometric limits
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 17 / 45
33. Limits and arithmetic
Fact
Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then
x→a x→a
1. lim [f (x) + g (x)] = L + M
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 18 / 45
34. Limits and arithmetic
Fact
Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then
x→a x→a
1. lim [f (x) + g (x)] = L + M (errors add)
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 18 / 45
35. Limits and arithmetic
Fact
Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then
x→a x→a
1. lim [f (x) + g (x)] = L + M (errors add)
x→a
2. lim [f (x) − g (x)] = L − M
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 19 / 45
36. Limits and arithmetic
Fact
Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then
x→a x→a
1. lim [f (x) + g (x)] = L + M (errors add)
x→a
2. lim [f (x) − g (x)] = L − M
x→a
3. lim [cf (x)] = cL
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 19 / 45
37. Limits and arithmetic
Fact
Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then
x→a x→a
1. lim [f (x) + g (x)] = L + M (errors add)
x→a
2. lim [f (x) − g (x)] = L − M
x→a
3. lim [cf (x)] = cL (error scales)
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 19 / 45
38. Justification of the scaling law
errors scale: If f (x) is e away from L, then
(c · f (x) − c · L) = c · (f (x) − L) = c · e
That is, (c · f )(x) is c · e away from cL,
So if Emerson gives us an error of 1 (for instance), Dana can use the
fact that lim f (x) = L to find a tolerance for f and g corresponding
x→a
to the error 1/c.
Dana wins the round.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 20 / 45
39. Limits and arithmetic
Fact
Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then
x→a x→a
1. lim [f (x) + g (x)] = L + M (errors add)
x→a
2. lim [f (x) − g (x)] = L − M (combination of adding and scaling)
x→a
3. lim [cf (x)] = cL (error scales)
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 21 / 45
40. Limits and arithmetic
Fact
Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then
x→a x→a
1. lim [f (x) + g (x)] = L + M (errors add)
x→a
2. lim [f (x) − g (x)] = L − M (combination of adding and scaling)
x→a
3. lim [cf (x)] = cL (error scales)
x→a
4. lim [f (x)g (x)] = L · M
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 22 / 45
41. Limits and arithmetic
Fact
Suppose lim f (x) = L and lim g (x) = M and c is a constant. Then
x→a x→a
1. lim [f (x) + g (x)] = L + M (errors add)
x→a
2. lim [f (x) − g (x)] = L − M (combination of adding and scaling)
x→a
3. lim [cf (x)] = cL (error scales)
x→a
4. lim [f (x)g (x)] = L · M (more complicated, but doable)
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 22 / 45
42. Limits and arithmetic II
Fact (Continued)
f (x) L
5. lim = , if M = 0.
x→a g (x) M
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 23 / 45
43. Caution!
The quotient rule for limits says that if lim g (x) = 0, then
x→a
f (x) limx→a f (x)
lim =
x→a g (x) limx→a g (x)
It does NOT say that if lim g (x) = 0, then
x→a
f (x)
lim does not exist
x→a g (x)
In fact, limits of quotients where numerator and denominator both
tend to 0 are exactly where the magic happens.
more about this later
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 24 / 45
44. Limits and arithmetic II
Fact (Continued)
f (x) L
5. lim = , if M = 0.
x→a g (x) M
n
6. lim [f (x)]n = lim f (x)
x→a x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
45. Limits and arithmetic II
Fact (Continued)
f (x) L
5. lim = , if M = 0.
x→a g (x) M
n
6. lim [f (x)]n = lim f (x) (follows from 4 repeatedly)
x→a x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
46. Limits and arithmetic II
Fact (Continued)
f (x) L
5. lim = , if M = 0.
x→a g (x) M
n
6. lim [f (x)]n = lim f (x) (follows from 4 repeatedly)
x→a x→a
7. lim x n = an
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
47. Limits and arithmetic II
Fact (Continued)
f (x) L
5. lim = , if M = 0.
x→a g (x) M
n
6. lim [f (x)]n = lim f (x) (follows from 4 repeatedly)
x→a x→a
7. lim x n = an
x→a
√ √
8. lim n x = n a
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
48. Limits and arithmetic II
Fact (Continued)
f (x) L
5. lim = , if M = 0.
x→a g (x) M
n
6. lim [f (x)]n = lim f (x) (follows from 4 repeatedly)
x→a x→a
7. lim x n = an (follows from 6)
x→a
√ √
8. lim n x = n a
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
49. Limits and arithmetic II
Fact (Continued)
f (x) L
5. lim = , if M = 0.
x→a g (x) M
n
6. lim [f (x)]n = lim f (x) (follows from 4 repeatedly)
x→a x→a
7. lim x n = an (follows from 6)
x→a
√ √
8. lim n x = n a
x→a
n
9. lim f (x) = n lim f (x) (If n is even, we must additionally assume
x→a x→a
that lim f (x) > 0)
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 25 / 45
50. Applying the limit laws
Example
Find lim x 2 + 2x + 4 .
x→3
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
51. Applying the limit laws
Example
Find lim x 2 + 2x + 4 .
x→3
Solution
By applying the limit laws repeatedly:
lim x 2 + 2x + 4
x→3
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
52. Applying the limit laws
Example
Find lim x 2 + 2x + 4 .
x→3
Solution
By applying the limit laws repeatedly:
lim x 2 + 2x + 4 = lim x 2 + lim (2x) + lim (4)
x→3 x→3 x→3 x→3
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
53. Applying the limit laws
Example
Find lim x 2 + 2x + 4 .
x→3
Solution
By applying the limit laws repeatedly:
lim x 2 + 2x + 4 = lim x 2 + lim (2x) + lim (4)
x→3 x→3 x→3 x→3
2
= lim x + 2 · lim (x) + 4
x→3 x→3
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
54. Applying the limit laws
Example
Find lim x 2 + 2x + 4 .
x→3
Solution
By applying the limit laws repeatedly:
lim x 2 + 2x + 4 = lim x 2 + lim (2x) + lim (4)
x→3 x→3 x→3 x→3
2
= lim x + 2 · lim (x) + 4
x→3 x→3
2
= (3) + 2 · 3 + 4
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
55. Applying the limit laws
Example
Find lim x 2 + 2x + 4 .
x→3
Solution
By applying the limit laws repeatedly:
lim x 2 + 2x + 4 = lim x 2 + lim (2x) + lim (4)
x→3 x→3 x→3 x→3
2
= lim x + 2 · lim (x) + 4
x→3 x→3
2
= (3) + 2 · 3 + 4
= 9 + 6 + 4 = 19.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 26 / 45
56. Your turn
Example
x 2 + 2x + 4
Find lim
x→3 x 3 + 11
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 27 / 45
57. Your turn
Example
x 2 + 2x + 4
Find lim
x→3 x 3 + 11
Solution
19 1
The answer is = .
38 2
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 27 / 45
58. Direct Substitution Property
Theorem (The Direct Substitution Property)
If f is a polynomial or a rational function and a is in the domain of f , then
lim f (x) = f (a)
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 28 / 45
59. Outline
Recall: The concept of limit
Basic Limits
Limit Laws
The direct substitution property
Limits with Algebra
Two more limit theorems
Two important trigonometric limits
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 29 / 45
60. Limits do not see the point! (in a good way)
Theorem
If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L.
x→a x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 30 / 45
61. Limits do not see the point! (in a good way)
Theorem
If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L.
x→a x→a
Example
x 2 + 2x + 1
Find lim , if it exists.
x→−1 x +1
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 30 / 45
62. Limits do not see the point! (in a good way)
Theorem
If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L.
x→a x→a
Example
x 2 + 2x + 1
Find lim , if it exists.
x→−1 x +1
Solution
x 2 + 2x + 1
Since = x + 1 whenever x = −1, and since lim x + 1 = 0,
x +1 x→−1
x 2 + 2x + 1
we have lim = 0.
x→−1 x +1
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 30 / 45
63. x 2 + 2x + 1
ET game for f (x) =
x +1
y
x
−1
Even if f (−1) were something else, it would not effect the limit.
64. x 2 + 2x + 1
ET game for f (x) =
x +1
y
x
−1
Even if f (−1) were something else, it would not effect the limit.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 31 / 45
65. Limit of a function defined piecewise at a boundary
point
Example
Let
x2 x ≥0
f (x) =
−x x <0
Does lim f (x) exist?
x→0
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
66. Limit of a function defined piecewise at a boundary
point
Example
Let
x2 x ≥0
f (x) =
−x x <0
Does lim f (x) exist?
x→0
Solution
We have
MTP DSP
lim+ f (x) = lim+ x 2 = 02 = 0
x→0 x→0
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
67. Limit of a function defined piecewise at a boundary
point
Example
Let
x2 x ≥0
f (x) =
−x x <0
Does lim f (x) exist?
x→0
Solution
We have
MTP DSP
lim+ f (x) = lim+ x 2 = 02 = 0
x→0 x→0
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
68. Limit of a function defined piecewise at a boundary
point
Example
Let
x2 x ≥0
f (x) =
−x x <0
Does lim f (x) exist?
x→0
Solution
We have
MTP DSP
lim+ f (x) = lim+ x 2 = 02 = 0
x→0 x→0
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
69. Limit of a function defined piecewise at a boundary
point
Example
Let
x2 x ≥0
f (x) =
−x x <0
Does lim f (x) exist?
x→0
Solution
We have
MTP DSP
lim+ f (x) = lim+ x 2 = 02 = 0
x→0 x→0
Likewise:
lim f (x) = lim −x = −0 = 0
x→0− x→0−
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
70. Limit of a function defined piecewise at a boundary
point
Example
Let
x2 x ≥0
f (x) =
−x x <0
Does lim f (x) exist?
x→0
Solution
We have
MTP DSP
lim+ f (x) = lim+ x 2 = 02 = 0
x→0 x→0
Likewise:
lim f (x) = lim −x = −0 = 0
x→0− x→0−
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
71. Limit of a function defined piecewise at a boundary
point
Example
Let
x2 x ≥0
f (x) =
−x x <0
Does lim f (x) exist?
x→0
Solution
We have
MTP DSP
lim+ f (x) = lim+ x 2 = 02 = 0
x→0 x→0
Likewise:
lim f (x) = lim −x = −0 = 0
x→0− x→0−
So lim f (x) = 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 32 / 45
72. Finding limits by algebraic manipulations
Example
√
x −2
Find lim .
x→4 x −4
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 33 / 45
73. Finding limits by algebraic manipulations
Example
√
x −2
Find lim .
x→4 x −4
Solution
√ 2 √ √
Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2).
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 33 / 45
74. Finding limits by algebraic manipulations
Example
√
x −2
Find lim .
x→4 x −4
Solution
√ 2 √ √
Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). So
√ √
x −2 x −2
lim = lim √ √
x→4 x − 4 x→4 ( x − 2)( x + 2)
1 1
= lim √ =
x→4 x +2 4
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 33 / 45
75. Your turn
Example
Let
1 − x2 x ≥1
f (x) =
2x x <1
Find lim f (x) if it exists.
x→1
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
76. Your turn
Example
Let
1 − x2 x ≥1
f (x) =
2x x <1
Find lim f (x) if it exists.
x→1
Solution
We have
DSP
lim+ f (x) = lim+ 1 − x 2 = 0
x→1 x→1
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
77. Your turn
Example
Let
1 − x2 x ≥1
f (x) =
2x x <1
Find lim f (x) if it exists. 1
x→1
Solution
We have
DSP
lim+ f (x) = lim+ 1 − x 2 = 0
x→1 x→1
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
78. Your turn
Example
Let
1 − x2 x ≥1
f (x) =
2x x <1
Find lim f (x) if it exists. 1
x→1
Solution
We have
DSP
lim+ f (x) = lim+ 1 − x 2 = 0
x→1 x→1
DSP
lim f (x) = lim (2x) = 2
x→1− x→1−
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
79. Your turn
Example
Let
1 − x2 x ≥1
f (x) =
2x x <1
Find lim f (x) if it exists. 1
x→1
Solution
We have
DSP
lim+ f (x) = lim+ 1 − x 2 = 0
x→1 x→1
DSP
lim f (x) = lim (2x) = 2
x→1− x→1−
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
80. Your turn
Example
Let
1 − x2 x ≥1
f (x) =
2x x <1
Find lim f (x) if it exists. 1
x→1
Solution
We have
DSP
lim+ f (x) = lim+ 1 − x 2 = 0
x→1 x→1
DSP
lim f (x) = lim (2x) = 2
x→1− x→1−
The left- and right-hand limits disagree, so the limit does not exist.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 34 / 45
81. A message from the Mathematical Grammar Police
Please do not say “ lim f (x) = DNE.” Does not compute.
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 35 / 45
82. A message from the Mathematical Grammar Police
Please do not say “ lim f (x) = DNE.” Does not compute.
x→a
Too many verbs
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 35 / 45
83. A message from the Mathematical Grammar Police
Please do not say “ lim f (x) = DNE.” Does not compute.
x→a
Too many verbs
Leads to FALSE limit laws like “If lim f (x) DNE and lim g (x) DNE,
x→a x→a
then lim (f (x) + g (x)) DNE.”
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 35 / 45
84. Two More Important Limit Theorems
Theorem
If f (x) ≤ g (x) when x is near a (except possibly at a), then
lim f (x) ≤ lim g (x)
x→a x→a
(as usual, provided these limits exist).
Theorem (The Squeeze/Sandwich/Pinching Theorem)
If f (x) ≤ g (x) ≤ h(x) when x is near a (as usual, except possibly at a),
and
lim f (x) = lim h(x) = L,
x→a x→a
then
lim g (x) = L.
x→a
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 36 / 45
85. Using the Squeeze Theorem
We can use the Squeeze Theorem to replace complicated expressions with
simple ones when taking the limit.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 37 / 45
86. Using the Squeeze Theorem
We can use the Squeeze Theorem to replace complicated expressions with
simple ones when taking the limit.
Example
π
Show that lim x 2 sin = 0.
x→0 x
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 37 / 45
87. Using the Squeeze Theorem
We can use the Squeeze Theorem to replace complicated expressions with
simple ones when taking the limit.
Example
π
Show that lim x 2 sin = 0.
x→0 x
Solution
We have for all x,
π π
−1 ≤ sin ≤ 1 =⇒ −x 2 ≤ x 2 sin ≤ x2
x x
The left and right sides go to zero as x → 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 37 / 45
88. Illustration of the Squeeze Theorem
y h(x) = x 2
x
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 38 / 45
89. Illustration of the Squeeze Theorem
y h(x) = x 2
x
f (x) = −x 2
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 38 / 45
90. Illustration of the Squeeze Theorem
y h(x) = x 2
x
π
2
g (x) = x sin
x
f (x) = −x 2
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 38 / 45
91. Outline
Recall: The concept of limit
Basic Limits
Limit Laws
The direct substitution property
Limits with Algebra
Two more limit theorems
Two important trigonometric limits
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 39 / 45
92. Two important trigonometric limits
Theorem
The following two limits hold:
sin θ
lim =1
θ→0 θ
cos θ − 1
lim =0
θ→0 θ
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 40 / 45
93. Proof of the Sine Limit
Proof.
Notice
θ
θ
θ
1
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
94. Proof of the Sine Limit
Proof.
Notice
sin θ ≤ θ
sin θ θ
θ
cos θ 1
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
95. Proof of the Sine Limit
Proof.
Notice
sin θ ≤ θ tan θ
sin θ θ tan θ
θ
cos θ 1
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
96. Proof of the Sine Limit
Proof.
Notice
θ
sin θ ≤ θ ≤ 2 tan ≤ tan θ
2
sin θ θ tan θ
θ
cos θ 1
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
97. Proof of the Sine Limit
Proof.
Notice
θ
sin θ ≤ θ ≤ 2 tan ≤ tan θ
2
Divide by sin θ:
θ 1
1≤ ≤
sin θ cos θ
sin θ θ tan θ
θ
cos θ 1
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
98. Proof of the Sine Limit
Proof.
Notice
θ
sin θ ≤ θ ≤ 2 tan ≤ tan θ
2
Divide by sin θ:
θ 1
1≤ ≤
sin θ cos θ
sin θ θ tan θ
θ Take reciprocals:
cos θ 1 sin θ
1≥ ≥ cos θ
θ
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
99. Proof of the Sine Limit
Proof.
Notice
θ
sin θ ≤ θ ≤ 2 tan ≤ tan θ
2
Divide by sin θ:
θ 1
1≤ ≤
sin θ cos θ
sin θ θ tan θ
θ Take reciprocals:
cos θ 1 sin θ
1≥ ≥ cos θ
θ
As θ → 0, the left and right sides tend to 1. So, then, must the middle
expression.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 41 / 45
100. Proof of the Cosine Limit
Proof.
1 − cos θ 1 − cos θ 1 + cos θ 1 − cos2 θ
= · =
θ θ 1 + cos θ θ(1 + cos θ)
2
sin θ sin θ sin θ
= = ·
θ(1 + cos θ) θ 1 + cos θ
So
1 − cos θ sin θ sin θ
lim = lim · lim
θ→0 θ θ→0 θ θ→0 1 + cos θ
= 1 · 0 = 0.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 42 / 45
101. Try these
Example
tan θ
1. lim
θ→0 θ
sin 2θ
2. lim
θ→0 θ
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 43 / 45
102. Try these
Example
tan θ
1. lim
θ→0 θ
sin 2θ
2. lim
θ→0 θ
Answer
1. 1
2. 2
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 43 / 45
103. Solutions
1. Use the basic trigonometric limit and the definition of tangent.
tan θ sin θ sin θ 1 1
lim = lim = lim · lim = 1 · = 1.
θ→0 θ θ→0 θ cos θ θ→0 θ θ→0 cos θ 1
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 44 / 45
104. Solutions
1. Use the basic trigonometric limit and the definition of tangent.
tan θ sin θ sin θ 1 1
lim = lim = lim · lim = 1 · = 1.
θ→0 θ θ→0 θ cos θ θ→0 θ θ→0 cos θ 1
2. Change the variable:
sin 2θ sin 2θ sin 2θ
lim = lim 1
= 2 · lim =2·1=2
θ→0 θ 2θ→0 2θ ·
2
2θ→0 2θ
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 44 / 45
105. Solutions
1. Use the basic trigonometric limit and the definition of tangent.
tan θ sin θ sin θ 1 1
lim = lim = lim · lim = 1 · = 1.
θ→0 θ θ→0 θ cos θ θ→0 θ θ→0 cos θ 1
2. Change the variable:
sin 2θ sin 2θ sin 2θ
lim = lim 1
= 2 · lim =2·1=2
θ→0 θ 2θ→0 2θ ·
2
2θ→0 2θ
OR use a trigonometric identity:
sin 2θ 2 sin θ cos θ sin θ
lim = lim = 2 · lim · lim cos θ = 2 · 1 · 1 = 2
θ→0 θ θ→0 θ θ→0 θ θ→0
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 44 / 45
106. Summary
The limit laws allow us to compute limits reasonably.
BUT we cannot make up extra laws otherwise we get into trouble.
V63.0121.021, Calculus I (NYU) Section 1.4 Calculating Limits September 16, 2010 45 / 45