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FUNCTIONS,
LIMITS AN
D CONTINUI
TY
1.Limits: Intuitive and Formal
Definition 2.Theorems on Limits
3.One-sided
Limits
•
•
•
Limits is our best prediction of a point we
didn’t observe.
Limits give us an estimate when we can’t
compute a result directly.
The limit wonders, “If you can see
everything except a single value, what do you
think is there?”.
1:07
1:08
Buffering…
1:09
1:10
1:11
LIMITS: INTUITIVE AND FORMAL
DEFINITION
1
x
y
y= f(x)
a
x 2.76 2.89 2.99 3.09 3.15 3.29 3.48
f(x) 6.52 6.78 6.98 7.18 7.30 7.58 7.96
f(x) =
2x+1
f(x) = 2x+1 is nearer
7
2�� + 1 → 7 ����
�� → 3
lim 2�� + 1
= 7
��→3
FORMAL DEFINITION OF LIMIT
Let f(x) be any function
Let a and L be the numbers
If we can make f(x) as close to L as we please
by choosing x sufficiently close to a then we
say that the limit of f(x) as x approaches a is
L or symbolically,
lim �� �� = �
�
��→��
We will study how f(x) changes as the value of
x approaches c, in symbols �� → ��.
Examples:
1. Consider �� �� =
��2 + 5 as x → 2
x 1.84 1.98 2.03 2.08 2.19 2.55
f(x) 8.39 8.92 9.12 9.33 9.80 11.50
��2 + 5 → 9 ��
�� �� → 2
lim ��2 + 5
= 9
��→2
2. �
�
��
2
�
� −
1
= 2�� +��−3
as ��
→ 1
x 0.53 0.72 0.89 0.99 1.04 1.12 1.34
f(x) 4.06 4.44 4.78 4.98 5.08 5.24 5.68
�
� −
1
2��2+��−3 → 5 �
��� �� → 1
lim
�
�
→1
2��2 +
�� − 3
��
− 1
= 5
3. Consider the piece-wise
function
�
�
�
�
��2 − 1 �
��� �� <
2
= { �� + 3 ��
�� �� ≥ 2
��
→ 2
��2 − 1 ��
�� �� < 2
x 1.85 1.89 1.999 1.9999
f(x) 2.42 2.57 2.996 2.9996
��
→2−
lim ��2 − 1
= 3
�� + 3 ��
�� �� ≥
2
x 2.1 2.01 2.001 2.0001
f(x) 5.1 5.01 5.001 5.0001
��
→2+
lim �� + 3
= 5
��
→2−
��
→2+
lim ��2 − 1 ≠
lim �� + 3
∴ lim ��
��
����
The limit of a function f(x) is equal to L
as x approaches a, written as
��
→�
�−
��
→�
�+
lim �� �� = �� if
and only if
��→��
lim �� �� = �
� = lim �� �
�
THEOREMS ON LIMITS
2
Let a and c be real numbers so
that
lim ��(��) and lim ��(��)
exist.
��→����→��
1. Constant Rule
Examples:
1.1 lim 8 = 8
��→��
1.2 lim 7.21 = 7.21
��→��
lim �� =
��
��→��
2. Identity
Rule
Examples:
2.1 lim �� =
5
��→5
2.2 lim ��
= 12
��→12
lim �� =
��
��→��
3. Sum and Difference Rule
lim[�� �� ± �� ��
= lim ��(��) ± lim ��(��)
��→�� ��→�� ��→��
Examples:
lim ��
��
= 5
��→��
3.1. lim
[�� �
�
��→��
+ ��
(��)
lim �� ��
= −8
��→��
+ �� �� ]
= lim �� ��
��→�
�
= 5 + (-8)
= -3
3.2. lim
[�� �
�
��→��
− ��
(��)
− �� ��
] = lim ��
��
��→�
�
= 5 - (-8)
4. Constant Multiple
Rule lim ��
��
�
�
��→��
= �� lim
��(��)
��→�
� = 7, then
Examples: If lim
�� ��
��→�
�
4.1.
lim −5 �
� ��
��→��
= −5 lim �
� ��
��→�
�
= −5
4
= −20
4.2. lim 4 �� �� = 4 lim
�� �� = 4 4 = 16
��→�� ��→��
5. Product
Rule
lim ��
��
⋅ ��
��
��→��
= lim ��(��) ∙
lim ��(��)
��→�� ��→��
Example:
Let lim �
� ��
��→��
= −12 and lim ��
�� = 3
��→��
�
�
→
�
�
lim �� �� ⋅ �� �
� = lim ��(��) ∙ lim ��
��
��→��
��
→��
= -12(3)
= -36
�
�
→
�
�
6. Quotient Rule, where lim ��(�
�) ≠ 0
lim
�
�(
�
�)
��→�
� ��
(��)
= ��→��
lim �
�(��)
lim �
�(��)
��→�
�
Examples:
6.1. If lim
�� ��
��→�
�
= 2 and lim �� �
� = −5
��→��
lim
�
�(
�
�)
��→�
� ��
(��)
lim ��
(��)
�
� →
�
�
= ��→��
=
2
lim ��(��)
−5
= −
2
5
6.2. If lim �� �� = 0 and lim
�� �� = −2
��→�� ��→��
lim
�
�(
�
�)
��→�
� ��
(��)
lim ��
(��)
�
� →
�
�
= ��→��
=
0
lim ��(��)
−2
= 0
6.3. If lim �� �� = 5 and
lim �� �� = 0
��→�� ��→��
DNE
not possible
to evaluate
7. Power
Rule
If n is a positive integer and f(x)≥ 0 if
n is even, then lim[�� �� ]�� =
[lim ��(��)]��
��→�� ��→��
Examples:
7.1. If lim
�� ��
��→�
�
= 3, then
lim(�� �� )3 = (lim ��
�� )3 = 33 = 27
��→�� ��→��
7.2. If lim �
�
��
��→�
�
= 3, then
��→�� ��→��
32
lim(�� �� )−2 = (lim �� �� )−2
= 3−2 = 1 = 1 9
8. Root
Rule
If n is a positive integer and f(x)≥ 0 if n
is even, then lim � � ��(��) = � �
lim ��(��)
��→�� ��→��
= 9, then
Examples:
8.1. If lim
����
��→��
lim
� �
��
→�
�
��(�
�) =
� �
lim ��(��) =
9 = 3
��→��
�
�
→
�
�
8.2. If lim �� �� = −9, then it is
not possible to
evaluate because
lim ��(��)
= −9
��→��
Examples:
1. Find lim(2��2 + 3�� + 4)
��→2
Solution:
lim(2��2 + 3�� + 4) = lim 2��2 + lim
3�� + lim 4
��→2 ��→2 ��→2 ��→2
= 2 lim ��2 + 3 lim ��
+ lim 4
��→2 ��→2 ��
→2
= 2 (lim ��)2 + 3 lim ��
+ lim 4
� � →2
� �
→2
+ 4
��→2
= 2(2)2 + 32
= 2(4) + 6 +
4
= 8 + 6 + 4
Sum and Difference
Rules
Constant Multiple Rule
Power Rule
Identity and
Constant Rules
∴ ������(���
��� + ���� + �
2. Evaluate lim
��
2+2��
−3
�
�+
7
��→−2
Solution:
lim
� �
→−2
��2 +
2�� − 3
��
+ 7
= ��→−2 ��→−2 ��
→−2
lim ��2 + lim
2�� − lim 3
� � →−2 �
� →−2
lim �� +
lim 7
= ��→−2 ��→−2 ��→−2
( lim ��)2+2 lim
��− lim 3
� � →−2
� �
→−2
lim � � +
lim 7
2
= (−2) +2 −2 −3
−2+7
= 4−4−3
5
= - 3
5
��
→−�
�
��
∴ �����
� �� +����
−��
=-��
��+��
��
3. Evaluate lim(�
� + 4)
��→2
2��
+ 5
Solution:
lim(��
+ 4)
��→2
�
�
→2
�
�
→2
2�� + 5 = (lim �� +
lim 4)
lim(2��
+ 5)
��→2
�
�
→2
= (lim �� +
lim 4)
� � →2 � � →2
� � →2
lim 2�� +
lim 5
= (2 + 4) 2 lim �� +
lim 5
��→2 �
�→2
= 6 [ 2 2
+ 5]
= 6 ( 4 + 5)
= 6 ( 9)
= 6 (3)
= 18
∴ ����
��(�� +
��)
����
+ ��=18
�
�
→0
4. lim tan(�
�) �
�
Solution:
�
�
→0
�
�
�
�
→0
sin
�
�
tan(��)
cos�� lim =
lim
=
lim
�
�
sin x
��→0 �
� cos ��
�
�
→0
= lim
sin x ⋅
1
�� cos
��
= 1 ∙
lim
1
��→0
cos ��
= 1 ⋅
1
1
= 1
Note:
sin
�
�
tan �� =
cos
��
Quotient Identity
Substitute the above identity to ��
���� x
Simplify
Separate the limit using the product rule
�
�→
0
�
�
Applying the lim sin x =
1
Substitute x = 0 in ���
��� x
∴ lim
�
�
tan(
� � )
�
�
=1
Try!
5. lim(3��
+ 4)2
��→3
�
�
→3
5. lim(3�� +
4)2
Solution:
lim(3�� + 4)2 = [lim(3�
� + 4)]2
��→3 ��→3
= [lim 3�� + lim
4]2
��→3 ��→3
�
�
→3
= [3 lim �� + lim
4]2
+ 4]2
�
�
→3
= [3 3
= 169
Using the limit laws, evaluation of limits
of functions is most of the time tedious.
But there is an easier way to do so.
DIRECT SUBSTITUTION
Examples:
�
�
→2
1. Evaluate lim(2��2 + 3�� + 4)
��→2
Solution:
lim(2��2 + 3�� + 4) = 2(2)2 + 3
2 + 4
��→2
= 2(4) + 6 + 4
= 8 + 6 + 4
= 18
∴ lim(2��2 + 3�� + 4)=18
Some limits cannot be evaluated simply using
direct substitution. Particularly in cases when
the given is a rational function, direct
substitution sometimes yields a number where
both the numerator and denominator are 0. The
expressio
n
0
0 is an example of an indeterminate
form.
2. Evaluate lim �
�−2 ��→2 �
�3−8
Solution:
lim
��
− 2
��
→2
��
3
= lim
��
− 2
− 8 ��→2 (�� −
2)(��
2 + 2�� +
4)
= lim
1
��→2 ��
2+2��+4
1
=
=
(2)2+2 2 +4
1
4+4+4
= 1
12 ∴ lim �
�−2 =
1
��→2 ��3−8
12
�
�
→0
2
3. Evaluate lim (�
�+2) −4 �
�
S olution:
lim
�
�
→0
(�� + 2)2
− 4 �
�
=
��2 + 4��
+ 4 − 4�
�
2
= ��
+
4��
�
�
= ��(�
�+4)
�
�
�
�
��
= x + 4
lim �� + 4 = 0 + 4 = 4
��→0
(��+2)2−4
∴ lim =4
�
�
→5
2
4. lim ��
−10��
+25
��
2−25
5. li
m
�
�
→4
�
� −
4
�
� −
2
�
�
→5
2
4. lim ��
−10��+25
��
2−25
Solution:
��2 − 10�
� + 25
��2
− 25
(�� − 5)(�
� − 5)
=
(�� − 5)(�
� + 5)
��
− 5
��→5
�
� + 5
lim =
5 − 50
5 + 510
= = 0
�
�
→5
��
2−25
2
∴ lim ��
−10��+25
=0
5. li
m
�
�
→4
�
� −
4
�
�−
2
Solution:
�� − 4
��
− 2
=
��
− 4
��
− 2
⋅
��
+ 2
��
+ 2
=
�� − 4 �
� + 2
�� − 4
= �
� + 2
lim
�
�
→4
��
− 4
��
− 2
= lim
�
�
→4
�� +
2 =
4 + 2 = 2 + 2 = 4
∴ lim
�
�
→4
�
� −
2
��−4
=4
ONE-SIDED LIMITS
3
•


One-sided limits are the same as
normal limits, we just restrict x so that
it approaches from just one side.
Right-hand Limit
Left-hand Limit
RIGHT-HAND LIMIT
Let f be a fu n c ti on def ined on s om e open
interval. Then the limit of the function f as x
approaches a from the right is L, which is
written as
lim �� �� = ��
��→��+
LEFT-HAND LIMIT
Let f be a fu n c ti on def ined on s om e open
interval. Then the limit of the function f as x
approac h es a from th e left is L , wh ich is
written as
lim �� �� = ��
��→��−
�
�
→
�
�
The lim ��(��) exists and is
equal to L iff
i. lim ��(��) and lim �
�(��) exists; and
��→��−
��→��
+
ii. lim ��(��) = lim ��(�
�) = L
��→��−
��→��
+
Example: The graph of a function f is shown
below. Use it to state the values (if they exist) of
the following:
a. lim �� �� b.
��→2− ��
→2+
lim
��
��
c. lim
�� �
�
��→2
��
→5−
d. lim �� �� e.
��
→5+
lim
��
��
f.lim �
� �
�
��→5
Example: The graph of a function g is shown
below. Use it to state the values (if they exist) of
the following:
a. lim �� �� b. lim
�� ��
��→1− ��→1+
c. lim
�� �
�
��→1
d. lim �� �� e. g(5)
��→5
Example: Evaluate the following one-sided
limits:
lim 9 −
��
��→9−
lim 9 − �
�
��→9+
lim 9 − �� =
0
��→9−
lim 9 − �� = �
�����
��→9+
∴ lim
�
�
→9
9 − �� = �
�����.
Example: Let f be defined as
�� ��
4 − ��2
���� �
� ≤ 1
= {
2 + ��2 ��
�� �� > 1
Evaluate the following one-sided
limits
��
→1−
lim 4 −
��2 ��
→1+
lim 2 +
��2
��
→1−
lim 4 − 1 2
= 3
��
→1+
lim 2 + (1.0001)2 =
3
∴ lim ��
��
= 3
��→1
Try!
Determine if lim ��(�
�) exist.
��→3
�
�
�
�
= { ��
− 1��2
− 7
�
�
�
�
��
≤ 3
�� >
3
�
�
�
�
= { ��
− 1��2
− 7
����
��
≤ 3
�� >
3
lim �� − 1 = 3 − 1 =
2
��→3
∴ lim �� − 1 = 2
��→3
lim ��2 − 7 = (3.01)2 − 7 =
2
��→3
∴ lim ��2 − 7 = 2
��→3
∴ ��ℎ�� lim �� ��
���������� �
�ℎ����ℎ ���� 2.
��→3

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Limits of a function: Introductory to Calculus

  • 2. 1.Limits: Intuitive and Formal Definition 2.Theorems on Limits 3.One-sided Limits
  • 3. • • • Limits is our best prediction of a point we didn’t observe. Limits give us an estimate when we can’t compute a result directly. The limit wonders, “If you can see everything except a single value, what do you think is there?”.
  • 9. LIMITS: INTUITIVE AND FORMAL DEFINITION 1 x y y= f(x) a
  • 10. x 2.76 2.89 2.99 3.09 3.15 3.29 3.48 f(x) 6.52 6.78 6.98 7.18 7.30 7.58 7.96 f(x) = 2x+1 f(x) = 2x+1 is nearer 7 2�� + 1 → 7 ���� �� → 3 lim 2�� + 1 = 7 ��→3
  • 11. FORMAL DEFINITION OF LIMIT Let f(x) be any function Let a and L be the numbers If we can make f(x) as close to L as we please by choosing x sufficiently close to a then we say that the limit of f(x) as x approaches a is L or symbolically, lim �� �� = � � ��→��
  • 12. We will study how f(x) changes as the value of x approaches c, in symbols �� → ��. Examples: 1. Consider �� �� = ��2 + 5 as x → 2 x 1.84 1.98 2.03 2.08 2.19 2.55 f(x) 8.39 8.92 9.12 9.33 9.80 11.50 ��2 + 5 → 9 �� �� �� → 2 lim ��2 + 5 = 9 ��→2
  • 13. 2. � � �� 2 � � − 1 = 2�� +��−3 as �� → 1 x 0.53 0.72 0.89 0.99 1.04 1.12 1.34 f(x) 4.06 4.44 4.78 4.98 5.08 5.24 5.68 � � − 1 2��2+��−3 → 5 � ��� �� → 1 lim � � →1 2��2 + �� − 3 �� − 1 = 5
  • 14. 3. Consider the piece-wise function � � � � ��2 − 1 � ��� �� < 2 = { �� + 3 �� �� �� ≥ 2 �� → 2 ��2 − 1 �� �� �� < 2 x 1.85 1.89 1.999 1.9999 f(x) 2.42 2.57 2.996 2.9996 �� →2− lim ��2 − 1 = 3
  • 15. �� + 3 �� �� �� ≥ 2 x 2.1 2.01 2.001 2.0001 f(x) 5.1 5.01 5.001 5.0001 �� →2+ lim �� + 3 = 5 �� →2− �� →2+ lim ��2 − 1 ≠ lim �� + 3 ∴ lim �� �� ����
  • 16. The limit of a function f(x) is equal to L as x approaches a, written as �� →� �− �� →� �+ lim �� �� = �� if and only if ��→�� lim �� �� = � � = lim �� � �
  • 17. THEOREMS ON LIMITS 2 Let a and c be real numbers so that lim ��(��) and lim ��(��) exist. ��→����→�� 1. Constant Rule Examples: 1.1 lim 8 = 8 ��→�� 1.2 lim 7.21 = 7.21 ��→�� lim �� = �� ��→��
  • 18. 2. Identity Rule Examples: 2.1 lim �� = 5 ��→5 2.2 lim �� = 12 ��→12 lim �� = �� ��→�� 3. Sum and Difference Rule lim[�� �� ± �� �� = lim ��(��) ± lim ��(��) ��→�� ��→�� ��→��
  • 19. Examples: lim �� �� = 5 ��→�� 3.1. lim [�� � � ��→�� + �� (��) lim �� �� = −8 ��→�� + �� �� ] = lim �� �� ��→� � = 5 + (-8) = -3 3.2. lim [�� � � ��→�� − �� (��) − �� �� ] = lim �� �� ��→� � = 5 - (-8)
  • 20. 4. Constant Multiple Rule lim �� �� � � ��→�� = �� lim ��(��) ��→� � = 7, then Examples: If lim �� �� ��→� � 4.1. lim −5 � � �� ��→�� = −5 lim � � �� ��→� � = −5 4 = −20 4.2. lim 4 �� �� = 4 lim �� �� = 4 4 = 16 ��→�� ��→��
  • 21. 5. Product Rule lim �� �� ⋅ �� �� ��→�� = lim ��(��) ∙ lim ��(��) ��→�� ��→�� Example: Let lim � � �� ��→�� = −12 and lim �� �� = 3 ��→�� � � → � � lim �� �� ⋅ �� � � = lim ��(��) ∙ lim �� �� ��→�� �� →�� = -12(3) = -36
  • 22. � � → � � 6. Quotient Rule, where lim ��(� �) ≠ 0 lim � �( � �) ��→� � �� (��) = ��→�� lim � �(��) lim � �(��) ��→� � Examples: 6.1. If lim �� �� ��→� � = 2 and lim �� � � = −5 ��→�� lim � �( � �) ��→� � �� (��) lim �� (��) � � → � � = ��→�� = 2 lim ��(��) −5 = − 2 5
  • 23. 6.2. If lim �� �� = 0 and lim �� �� = −2 ��→�� ��→�� lim � �( � �) ��→� � �� (��) lim �� (��) � � → � � = ��→�� = 0 lim ��(��) −2 = 0 6.3. If lim �� �� = 5 and lim �� �� = 0 ��→�� ��→�� DNE not possible to evaluate
  • 24. 7. Power Rule If n is a positive integer and f(x)≥ 0 if n is even, then lim[�� �� ]�� = [lim ��(��)]�� ��→�� ��→�� Examples: 7.1. If lim �� �� ��→� � = 3, then lim(�� �� )3 = (lim �� �� )3 = 33 = 27 ��→�� ��→��
  • 25. 7.2. If lim � � �� ��→� � = 3, then ��→�� ��→�� 32 lim(�� �� )−2 = (lim �� �� )−2 = 3−2 = 1 = 1 9
  • 26. 8. Root Rule If n is a positive integer and f(x)≥ 0 if n is even, then lim � � ��(��) = � � lim ��(��) ��→�� ��→�� = 9, then Examples: 8.1. If lim ���� ��→�� lim � � �� →� � ��(� �) = � � lim ��(��) = 9 = 3 ��→��
  • 27. � � → � � 8.2. If lim �� �� = −9, then it is not possible to evaluate because lim ��(��) = −9 ��→��
  • 28. Examples: 1. Find lim(2��2 + 3�� + 4) ��→2 Solution: lim(2��2 + 3�� + 4) = lim 2��2 + lim 3�� + lim 4 ��→2 ��→2 ��→2 ��→2 = 2 lim ��2 + 3 lim �� + lim 4 ��→2 ��→2 �� →2 = 2 (lim ��)2 + 3 lim �� + lim 4 � � →2 � � →2 + 4 ��→2 = 2(2)2 + 32 = 2(4) + 6 + 4 = 8 + 6 + 4 Sum and Difference Rules Constant Multiple Rule Power Rule Identity and Constant Rules ∴ ������(��� ��� + ���� + �
  • 29. 2. Evaluate lim �� 2+2�� −3 � �+ 7 ��→−2 Solution: lim � � →−2 ��2 + 2�� − 3 �� + 7 = ��→−2 ��→−2 �� →−2 lim ��2 + lim 2�� − lim 3 � � →−2 � � →−2 lim �� + lim 7 = ��→−2 ��→−2 ��→−2 ( lim ��)2+2 lim ��− lim 3 � � →−2 � � →−2 lim � � + lim 7 2 = (−2) +2 −2 −3 −2+7 = 4−4−3 5 = - 3 5 �� →−� � �� ∴ ����� � �� +���� −�� =-�� ��+�� ��
  • 30. 3. Evaluate lim(� � + 4) ��→2 2�� + 5 Solution: lim(�� + 4) ��→2 � � →2 � � →2 2�� + 5 = (lim �� + lim 4) lim(2�� + 5) ��→2 � � →2 = (lim �� + lim 4) � � →2 � � →2 � � →2 lim 2�� + lim 5 = (2 + 4) 2 lim �� + lim 5 ��→2 � �→2 = 6 [ 2 2 + 5] = 6 ( 4 + 5) = 6 ( 9) = 6 (3) = 18 ∴ ���� ��(�� + ��) ���� + ��=18
  • 31. � � →0 4. lim tan(� �) � � Solution: � � →0 � � � � →0 sin � � tan(��) cos�� lim = lim = lim � � sin x ��→0 � � cos �� � � →0 = lim sin x ⋅ 1 �� cos �� = 1 ∙ lim 1 ��→0 cos �� = 1 ⋅ 1 1 = 1 Note: sin � � tan �� = cos �� Quotient Identity Substitute the above identity to �� ���� x Simplify Separate the limit using the product rule � �→ 0 � � Applying the lim sin x = 1 Substitute x = 0 in ��� ��� x ∴ lim � � tan( � � ) � � =1
  • 33. � � →3 5. lim(3�� + 4)2 Solution: lim(3�� + 4)2 = [lim(3� � + 4)]2 ��→3 ��→3 = [lim 3�� + lim 4]2 ��→3 ��→3 � � →3 = [3 lim �� + lim 4]2 + 4]2 � � →3 = [3 3 = 169
  • 34. Using the limit laws, evaluation of limits of functions is most of the time tedious. But there is an easier way to do so. DIRECT SUBSTITUTION
  • 35. Examples: � � →2 1. Evaluate lim(2��2 + 3�� + 4) ��→2 Solution: lim(2��2 + 3�� + 4) = 2(2)2 + 3 2 + 4 ��→2 = 2(4) + 6 + 4 = 8 + 6 + 4 = 18 ∴ lim(2��2 + 3�� + 4)=18
  • 36. Some limits cannot be evaluated simply using direct substitution. Particularly in cases when the given is a rational function, direct substitution sometimes yields a number where both the numerator and denominator are 0. The expressio n 0 0 is an example of an indeterminate form.
  • 37. 2. Evaluate lim � �−2 ��→2 � �3−8 Solution: lim �� − 2 �� →2 �� 3 = lim �� − 2 − 8 ��→2 (�� − 2)(�� 2 + 2�� + 4) = lim 1 ��→2 �� 2+2��+4 1 = = (2)2+2 2 +4 1 4+4+4 = 1 12 ∴ lim � �−2 = 1 ��→2 ��3−8 12
  • 38. � � →0 2 3. Evaluate lim (� �+2) −4 � � S olution: lim � � →0 (�� + 2)2 − 4 � � = ��2 + 4�� + 4 − 4� � 2 = �� + 4�� � � = ��(� �+4) � � � � �� = x + 4 lim �� + 4 = 0 + 4 = 4 ��→0 (��+2)2−4 ∴ lim =4
  • 39. � � →5 2 4. lim �� −10�� +25 �� 2−25 5. li m � � →4 � � − 4 � � − 2
  • 40. � � →5 2 4. lim �� −10��+25 �� 2−25 Solution: ��2 − 10� � + 25 ��2 − 25 (�� − 5)(� � − 5) = (�� − 5)(� � + 5) �� − 5 ��→5 � � + 5 lim = 5 − 50 5 + 510 = = 0 � � →5 �� 2−25 2 ∴ lim �� −10��+25 =0
  • 41. 5. li m � � →4 � � − 4 � �− 2 Solution: �� − 4 �� − 2 = �� − 4 �� − 2 ⋅ �� + 2 �� + 2 = �� − 4 � � + 2 �� − 4 = � � + 2 lim � � →4 �� − 4 �� − 2 = lim � � →4 �� + 2 = 4 + 2 = 2 + 2 = 4 ∴ lim � � →4 � � − 2 ��−4 =4
  • 42. ONE-SIDED LIMITS 3 •   One-sided limits are the same as normal limits, we just restrict x so that it approaches from just one side. Right-hand Limit Left-hand Limit
  • 43. RIGHT-HAND LIMIT Let f be a fu n c ti on def ined on s om e open interval. Then the limit of the function f as x approaches a from the right is L, which is written as lim �� �� = �� ��→��+
  • 44. LEFT-HAND LIMIT Let f be a fu n c ti on def ined on s om e open interval. Then the limit of the function f as x approac h es a from th e left is L , wh ich is written as lim �� �� = �� ��→��−
  • 45. � � → � � The lim ��(��) exists and is equal to L iff i. lim ��(��) and lim � �(��) exists; and ��→��− ��→�� + ii. lim ��(��) = lim ��(� �) = L ��→��− ��→�� +
  • 46. Example: The graph of a function f is shown below. Use it to state the values (if they exist) of the following: a. lim �� �� b. ��→2− �� →2+ lim �� �� c. lim �� � � ��→2 �� →5− d. lim �� �� e. �� →5+ lim �� �� f.lim � � � � ��→5
  • 47. Example: The graph of a function g is shown below. Use it to state the values (if they exist) of the following: a. lim �� �� b. lim �� �� ��→1− ��→1+ c. lim �� � � ��→1 d. lim �� �� e. g(5) ��→5
  • 48. Example: Evaluate the following one-sided limits: lim 9 − �� ��→9− lim 9 − � � ��→9+ lim 9 − �� = 0 ��→9− lim 9 − �� = � ����� ��→9+ ∴ lim � � →9 9 − �� = � �����.
  • 49. Example: Let f be defined as �� �� 4 − ��2 ���� � � ≤ 1 = { 2 + ��2 �� �� �� > 1 Evaluate the following one-sided limits �� →1− lim 4 − ��2 �� →1+ lim 2 + ��2 �� →1− lim 4 − 1 2 = 3 �� →1+ lim 2 + (1.0001)2 = 3 ∴ lim �� �� = 3 ��→1
  • 50. Try! Determine if lim ��(� �) exist. ��→3 � � � � = { �� − 1��2 − 7 � � � � �� ≤ 3 �� > 3
  • 51. � � � � = { �� − 1��2 − 7 ���� �� ≤ 3 �� > 3 lim �� − 1 = 3 − 1 = 2 ��→3 ∴ lim �� − 1 = 2 ��→3 lim ��2 − 7 = (3.01)2 − 7 = 2 ��→3 ∴ lim ��2 − 7 = 2 ��→3 ∴ ��ℎ�� lim �� �� ���������� � �ℎ����ℎ ���� 2. ��→3