Week 3 Course Notes - MATH 137 WEEK 3 NOTES PROF DOUG PARK...

MATH 137 WEEK 3 NOTES
PROF. DOUG PARK
1. Limit laws
Limit Laws . Suppose f ( x ) → L and g ( x ) → M , as x → a . Then
• Sum Law: f ( x ) + g ( x ) → L + M
• Difference Law: f ( x ) - g ( x ) → L - M
• Constant Multiple Law: cf ( x ) → cL , where c = constant.
• Product Law: f ( x ) · g ( x ) → L · M
• Quotient Law:
f ( x )
g ( x )
→
L
M
, as long as M 6 = 0.
• Composition Law: g ( f ( x )) → g ( L ), as long as g ( x ) is continuous at x = L .
We will prove the Sum Law using the precise definition of limit. Proofs of other
laws are found in Appendix F of Textbook.
Proof of Sum Law . We need to prove lim
x → a
( f ( x ) + g ( x )) = L + M if lim
x → a
f ( x ) = L
and lim
x → a
g ( x ) = M .
Given any
1
> 0, there is δ
1
> 0 such that
0 < | x - a | < δ
1
= ⇒ | f ( x ) - L | <
1
Given any
2
> 0, there is δ
2
> 0 such that
0 < | x - a | < δ
2
= ⇒ | g ( x ) - M | <
2
Given any > 0, let δ = min { δ
1
, δ
2
} . Then
0 < | x - a | < δ = ⇒ | f ( x ) - L | <
1
and | g ( x ) - M | <
2
= ⇒ | f ( x ) - L | + | g ( x ) - M | <
1
+
2
= ⇒ | ( f ( x ) - L ) + ( g ( x ) - M ) | ≤ | f ( x ) - L | + | g ( x ) - M | <
1
+
2
= ⇒ | ( f ( x ) + g ( x )) - ( L + M ) | <
1
+
2
We are done if
1
+
2
= . Thus we set
1
=
2
=
2
.
2. Computing limits of
0
0
form
Ex . Evaluate L = lim
x → a
x
3
- a
3
x - a
( a = constant).
Sol’n . If we plug in x = a , we get
0
0
form.
In such cases we try to cancel the common zero factor.
x
3
- a
3
= ( x - a )( x
2
+ ax + a
2
)
Date : September 28, 2009.
1

2 DOUG PARK
Thus
L = lim
x → a
x - a
x - a
· ( x
2
+ ax + a
2
)
= lim
x → a
( x
2
+ ax + a
2
) = a
2
+ a
2
+ a
2
= 3 a
2
Ex . Evaluate L = lim
x → 2
x -
√
3 x - 2
√
x + 2 - 2
.
Sol’n .
0
0
form so we cancel ( x - 2) terms.
The ( x - 2) terms are found when we multiply by conjugate expressions!
Hence multiply both top and bottom by ( x +
√
3 x - 2)(
√
x + 2 + 2).
We get
( x 2 - 3 x + 2)(
√
x + 2 + 2)
( x + 2 - 4)( x +
√
3 x - 2)
=
( x - 1)( x - 2)(
√
x + 2 + 2)
( x - 2)( x +
√
3 x - 2)
=
( x - 1)(
√
x + 2 + 2)
x +
√
3 x - 2
-→
(2 - 1)(
√
2 + 2 + 2)
2 +
√
6 - 2
=
4
4
= 1
3. One-sided limits
One-sided limits will usually be of
constant
0
form, or limits involving | | symbol.
Theorem . lim
x → a
f ( x ) = L ⇐⇒ lim
x → a -
f ( x ) = lim
x → a +
f ( x ) = L
Ex . Evaluate the following limits if they exist.
(i) lim
x → 1
1
( x - 1) 3
(ii) lim
x → 1
1
| x - 1 |
(iii) lim
x → 1
x - 1
| x - 1 |
Strategy . Note that
1
0 +
= + ∞ ,
1
0 -
= -∞ .
Sol’n . (i) lim
x → 1 +
1
( x - 1) 3
=
1
(1 + - 1) 3
=
1
(0 + ) 3
=
1
0 +
= + ∞
lim
x → 1 -
1
( x - 1) 3
=
1
(1 - - 1) 3
=
1
(0 - ) 3
=
1
0 -
= -∞
Hence the limit does not exist (DNE).
(ii) lim
x → 1 +
1
| x - 1 |
=
1
| 0 + |
=
1
0 +
= + ∞
lim
x → 1 -
1
| x - 1 |
=
1
| 0 - |
=
1
0 +
= + ∞
Hence the limit is + ∞ , but DNE as a real number.

MATH 137 3
(iii) If x → 1
+
, then x > 1, so x - 1 > 0. Thus
x - 1
| x - 1 |
=
x - 1
x - 1
= 1
lim
x → 1 +
x - 1
| x - 1 |
= lim
x → 1 +
1 = 1
If x → 1 - , then x < 1, so x - 1 < 0. Thus
x - 1
| x - 1 |
=
x - 1
- ( x - 1)
= - 1
lim
x → 1 -
x - 1
| x - 1 |
= lim
x →