Computing limits uses some basic facts, some rules about how limits work with respect to algebra, and something called the Sandwich Theorem!

1. Section 1.4
Calculating Limits
V63.0121, Calculus I
January 28–29, 2009
Announcements
Homework 1 is due today. Put it in the folder corresponding
to your name
Please move table before class.
ALEKS initial due Friday
Please redo get-to-know-you survey (extra credit).

2. Outline
Basic Limits
Limit Laws
The direct substitution property
Limits with Algebra
Two more limit theorems
Two important trigonometric limits

3. 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

4. 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.

5. ET game for f (x) = x
y
x

6. ET game for f (x) = x
y
x

7. ET game for f (x) = x
y
a
x
a

8. ET game for f (x) = x
y
a
x
a

9. ET game for f (x) = x
y
a
x
a

10. ET game for f (x) = x
y
a
x
a

11. ET game for f (x) = x
y
a
x
a
Setting error equal to tolerance works!

12. ET game for f (x) = c

13. ET game for f (x) = c
y
x

14. ET game for f (x) = c
y
x

15. ET game for f (x) = c
y
c
x
a

16. ET game for f (x) = c
y
c
x
a

17. ET game for f (x) = c
y
c
x
a

18. ET game for f (x) = c
y
c
x
a
any tolerance works!

19. 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.

20. Outline
Basic Limits
Limit Laws
The direct substitution property
Limits with Algebra
Two more limit theorems
Two important trigonometric limits

21. Limits and arithmetic
Fact
Suppose lim f (x) and lim g (x) exist and c is a constant. Then
x→a x→a
1. lim [f (x) + g (x)] = lim f (x) + lim g (x)
x→a x→a x→a
2. lim [f (x) − g (x)] = lim f (x) − lim g (x)
x→a x→a x→a
3. lim [cf (x)] = c lim f (x)
x→a x→a
4. lim [f (x)g (x)] = lim f (x) · lim g (x)
x→a x→a x→a

22. Limits and arithmetic II
Fact (Continued)
lim f (x)
f (x)
= x→a
5. lim , if lim g (x) = 0.
x→a g (x) lim g (x) x→a
x→a

23. Limits and arithmetic II
Fact (Continued)
lim f (x)
f (x)
= x→a
5. lim , if lim g (x) = 0.
x→a g (x) lim g (x) x→a
x→a
n
n
6. lim [f (x)] = lim f (x)
x→a x→a

24. Limits and arithmetic II
Fact (Continued)
lim f (x)
f (x)
= x→a
5. lim , if lim g (x) = 0.
x→a g (x) lim g (x) x→a
x→a
n
n
6. lim [f (x)] = lim f (x) (follows from 4 repeatedly)
x→a x→a

25. Limits and arithmetic II
Fact (Continued)
lim f (x)
f (x)
= x→a
5. lim , if lim g (x) = 0.
x→a g (x) lim g (x) x→a
x→a
n
n
6. lim [f (x)] = lim f (x) (follows from 4 repeatedly)
x→a x→a

26. Limits and arithmetic II
Fact (Continued)
lim f (x)
f (x)
= x→a
5. lim , if lim g (x) = 0.
x→a g (x) lim g (x) x→a
x→a
n
n
6. lim [f (x)] = 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

27. Limits and arithmetic II
Fact (Continued)
lim f (x)
f (x)
= x→a
5. lim , if lim g (x) = 0.
x→a g (x) lim g (x) x→a
x→a
n
n
6. lim [f (x)] = 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

28. Limits and arithmetic II
Fact (Continued)
lim f (x)
f (x)
= x→a
5. lim , if lim g (x) = 0.
x→a g (x) lim g (x) x→a
x→a
n
n
6. lim [f (x)] = 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) = lim f (x) (If n is even, we must additionally
n
x→a x→a
assume that lim f (x) > 0)
x→a

29. Applying the limit laws
Example
Find lim x 2 + 2x + 4 .
x→3

30. 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

31. 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

32. 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
+ 2 · lim (x) + 4
= lim x
x→3 x→3

33. 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
+ 2 · lim (x) + 4
= lim x
x→3 x→3
2
= (3) + 2 · 3 + 4

34. 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
+ 2 · lim (x) + 4
= lim x
x→3 x→3
2
= (3) + 2 · 3 + 4
= 9 + 6 + 4 = 19.

35. Your turn
Example
x 2 + 2x + 4
Find lim
x 3 + 11
x→3

36. Your turn
Example
x 2 + 2x + 4
Find lim
x 3 + 11
x→3
Solution
19 1
The answer is =.
38 2

37. 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

38. Outline
Basic Limits
Limit Laws
The direct substitution property
Limits with Algebra
Two more limit theorems
Two important trigonometric limits

39. 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

40. 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

41. 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
= x + 1 whenever x = −1, and since
Since
x +1
x 2 + 2x + 1
lim x + 1 = 0, we have lim = 0.
x +1
x→−1 x→−1

42. 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.

43. 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.

44. 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 x 2 = 02 = 0
lim f (x) =
x→0+ x→0+
Likewise:
lim f (x) = lim −x = −0 = 0
x→0− x→0−
So lim f (x) = 0.
x→0

45. 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 x 2 = 02 = 0
lim f (x) =
x→0+ x→0+
Likewise:
lim f (x) = lim −x = −0 = 0
x→0− x→0−
So lim f (x) = 0.
x→0

46. Finding limits by algebraic manipulations
Example √
x −2
Find lim .
x −4
x→4

47. 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).

48. 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 √ =
4
x +2
x→4

49. Your turn
Example
Let
1 − x2 x ≥1
f (x) =
2x x <1
Find lim f (x) if it exists.
x→1

50. 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
DSP
lim f (x) = lim (2x) = 2
x→1− x→1−
The left- and right-hand limits disagree, so the limit does not exist.

51. Your turn
Example
Let
1 − x2 x ≥1
f (x) =
2x x <1
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
DSP
lim f (x) = lim (2x) = 2
x→1− x→1−
The left- and right-hand limits disagree, so the limit does not exist.

52. 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

53. We can use the Squeeze Theorem to make complicated limits
simple.

54. We can use the Squeeze Theorem to make complicated limits
simple.
Example
π
Show that lim x 2 sin = 0.
x
x→0

55. We can use the Squeeze Theorem to make complicated limits
simple.
Example
π
Show that lim x 2 sin = 0.
x
x→0
Solution
We have for all x,
π
−x 2 ≤ x 2 sin ≤ x2
x
The left and right sides go to zero as x → 0.

56. Illustration of the Squeeze Theorem
h(x) = x 2
y
x

57. Illustration of the Squeeze Theorem
h(x) = x 2
y
x
f (x) = −x 2

58. Illustration of the Squeeze Theorem
h(x) = x 2
y
1
g (x) = x 2 sin
x
x
f (x) = −x 2

59. Outline
Basic Limits
Limit Laws
The direct substitution property
Limits with Algebra
Two more limit theorems
Two important trigonometric limits

60. Two important trigonometric limits
Theorem
The following two limits hold:
sin θ
lim =1
θ→0 θ
cos θ − 1
lim =0
θ
θ→0

61. 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.

62. Proof of the Cosine Limit
Proof.
1 − cos2 θ
1 − cos θ 1 − cos θ 1 + cos 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 θ→0 θ θ→0 1 + cos θ
θ→0
= 1 · 0 = 0.

63. Try these
Example
tan θ
lim
θ
θ→0
sin 2θ
lim
θ
θ→0

64. Try these
Example
tan θ
lim
θ
θ→0
sin 2θ
lim
θ
θ→0
Answer
1
2