𝑦
Continuity of Definition. Continuity of a Function
a Function
The function 𝑓 is said to be continuous at the number 𝑎 if
and only if the following three conditions are satisfied:
(I) 𝑓(𝑎) exists. If one or more of these conditions
(II) lim 𝑓 𝑥 exists. fails to hold at 𝑎, the function 𝑓 is said
𝑥→𝑎
(III) lim 𝑓 𝑥 = 𝑓(𝑎) to be “discontinuous”.
𝑥→𝑎
Example 1. A wholesaler sells a product by the pound, charges
$2 per pound if 10𝑙𝑏 or less are ordered. If more than, the
wholesaler charges $20 plus $1.40 for each pound excess of
10𝑙𝑏.
If we let 𝐶(𝑥) be the total cost at 𝑥 pound, construct the 5
appropriate piecewise function and determine if 𝐶 is
continuous at 𝑥 = 10. 𝑥
2𝑥 if 0 ≤ 𝑥 ≤ 10
Solution. 𝐶 𝑥 =ቊ
1.4𝑥 + 6 if 𝑥 > 10
Since 𝐶 10 = lim 𝐶 𝑥
In this function, 𝐶 10 = 20 𝑥→10
∴ 𝐶 is continuous at 𝑥 = 10.
lim− 𝐶(𝑥) = lim− 𝐶(𝑥) = 2𝑥 = 2(10) = 20
𝑥→10 𝑥→10
Basic Calculus
PUPSHS lim 𝐶(𝑥) = lim+ 𝐶(𝑥) = (1.4𝑥 + 6) = 1.4 10 + 6 = 20 Therefore, lim 𝐶 𝑥 = 20
𝑥→10+ 𝑥→10 𝑥→10
Continuity of
a Function Example 2. Find the value of 𝑘 that make the function
continuous everywhere.
3𝑥 + 7 if 𝑥 ≤ 4
𝑓 𝑥 =ቊ
𝑘𝑥 − 1 if 𝑥 > 4
Solution. First, let us use the three conditions of continuity
of a function
(I). 𝑓 4 = 3𝑥 + 7 = 3 4 + 7 = 19
lim 𝑓(𝑥) = lim− (3𝑥 + 7) = 3 4 + 7 = 19
𝑥→4− 𝑥→4
(II).
lim 𝑓(𝑥) = lim− 𝑘𝑥 − 1 = 4𝑘 − 1
𝑥→4+ 𝑥→4
Equate the RHL expression to 19 to force it to
be continuous.
That is, 4𝑘 − 1 = 19
4𝑘 = 19 + 1
4𝑘 = 20
𝑘=5
Basic Calculus
PUPSHS
Continuity of
𝑦
a Function Example 2. Find the value of 𝑘 that make the function
continuous everywhere.
3𝑥 + 7 if 𝑥 ≤ 4
𝑓 𝑥 =ቊ
𝑘𝑥 − 1 if 𝑥 > 4
Solution. First, let us use the three conditions of continuity
of a function
(I). 𝑓 4 = 3𝑥 + 7 = 3 4 + 7 = 19
lim 𝑓(𝑥) = lim− (3𝑥 + 7) = 3 4 + 7 = 19
𝑥→4− 𝑥→4
(II).
lim 𝑓(𝑥) = lim− 𝑘𝑥 − 1 = 4 5 − 1 = 19
𝑥→4+ 𝑥→4
Equate the RHL expression to 19 to force it to
be continuous.
That is, 4𝑘 − 1 = 19 Hence, lim 𝑓 𝑥 =
𝑥→4
4𝑘 = 19 + 1 19
4𝑘 = 20 (III). 𝑓 4 = lim 𝑓 𝑥 𝑥
𝑥→4
𝑘=5
Basic Calculus
PUPSHS ∴ At 𝑘 = 5, the function 𝑓(𝑥) is continuous everywhere.
Continuity of
a Function 𝑦
Example 3. Let 𝑔 be defined
by
2𝑥 + 3 if 𝑥 ≠ 1
𝑔 𝑥 =ቊ
2 if 𝑥 = 1
Determine if 𝑔 is continuous at 𝑥 =
1.
Solution. Investigating 𝑔 using the three conditions,
(I). 𝑔 1 = 2
lim 𝑔(𝑥) = lim−(2𝑥 + 3) = 2 1 + 3 = 5
𝑥→1− 𝑥→1
(II).
lim 𝑔(𝑥) = lim+(2𝑥 + 3) = 2 1 + 3 = 5
𝑥→1+ 𝑥→1
Thus, lim 𝑔 𝑥 = 5
𝑥→1
(III). 𝑔 1 ≠ lim 𝑔 𝑥
𝑥→1
Since conditions (I) and (II) are the only conditions
𝑥 satisfied, we can say that 𝑔 is discontinuous at 𝑥 = 1.
Basic Calculus
PUPSHS
Continuity of
a Function 𝑦
Example 3. Let 𝑔 be defined
by
2𝑥 + 3 if 𝑥 ≠ 1
Note: 𝑔 𝑥 =ቊ
2 if 𝑥 = 1
If 𝑔(1) were defined
to be 5, then Determine if 𝑔 is continuous at 𝑥 =
1.
Solution. Investigating 𝑔 using the three conditions,
𝑔 1 = lim 𝑔 𝑥
𝑥→1
(I). 𝑔 1 = 2
lim 𝑔(𝑥) = lim−(2𝑥 + 3) = 2 1 + 3 = 5
𝑥→1− 𝑥→1
(II).
lim 𝑔(𝑥) = lim+(2𝑥 + 3) = 2 1 + 3 = 5
𝑥→1+ 𝑥→1
Thus, lim 𝑔 𝑥 = 5
𝑥→1
(III). 𝑔 1 ≠ lim 𝑔 𝑥
𝑥→1
Since conditions (I) and (II) are the only conditions
𝑥 satisfied, we can say that 𝑔 is discontinuous at 𝑥 = 1.
Basic Calculus
PUPSHS
Continuity of
a Function 𝑦
Example 3. Let 𝑔 be defined
by
2𝑥 + 3 if 𝑥 ≠ 1
Note: 𝑔 𝑥 =ቊ
5 if 𝑥 = 1
If 𝑔(1) were defined
to be 5, then Determine if 𝑔 is continuous at 𝑥 =
1.
Solution. Investigating 𝑔 using the three conditions,
𝑔 1 = lim 𝑔 𝑥
𝑥→1
(I). 𝑔 1 = 5
and now continuous
at 𝑥 = 1
lim 𝑔(𝑥) = lim−(2𝑥 + 3) = 2 1 + 3 = 5
𝑥→1− 𝑥→1
(II).
lim 𝑔(𝑥) = lim+(2𝑥 + 3) = 2 1 + 3 = 5
𝑥→1+ 𝑥→1
Thus, lim 𝑔 𝑥 = 5
𝑥→1
(III). 𝑔 1 = lim 𝑔 𝑥
𝑥→1
Since all conditions are satisfied, we can say that 𝑔 is
continuous at 𝑥 = 1.
𝑥
This type of discontinuity is called removable
discontinuity. If it is not removable, we say it is essential
Basic Calculus
discontinuity.
PUPSHS
𝑦
Continuity of Example 5. Let 𝐹 be defined
a Function by 1
if 𝑥 ≠ 2 1
𝐹 𝑥 = ቐ𝑥 − 2 Determine if 𝐹 is ℎ 𝑥 =
𝑥−2
−1 continuous at 𝑥 =
if 𝑥 = 2
2.
Solution. Using the three conditions,
𝑥
(I). 𝐹 2 = −1
𝑥=2
1
lim− 𝐹(𝑥) = lim− = −∞
𝑥→2 𝑥→2 𝑥 − 2
(II). 1
lim+ 𝐹(𝑥) = 𝑥→2
lim+
𝑥 − 2 = +∞
𝑥→2
1
Thus, lim 𝐹(𝑥) does not Example 4. Let ℎ be definedℎ 𝑥 =
𝑥→2
by 𝑥−2
exist.
Since (II) is not satisfied, we can conclude that 𝐹 Determine if ℎ is continuous at 𝑥 =
has discontinuity at 𝑥 = 2. 2.
Solution. Using the three conditions,
𝑦
1 (I). ℎ 2 is not defined.
ℎ 𝑥 =
𝑥−2
𝑥=2
Since (I) is not satisfied, we can conclude that ℎ
has discontinuity at 𝑥 = 2.
Example 5 is of𝑥 course an This type of discontinuity is called infinite
infinite discontinuity, at discontinuity.
Basic Calculus
the same time, essential
PUPSHS
discontinuity.
Continuity of Example 6. Let 𝐺 be defined
a Function 𝑦
by
3 + 𝑥 if 𝑥 ≤ 1
𝐺 𝑥 =ቊ
3 − 𝑥 if 𝑥 > 1
Investigate the discontinuity at 𝑥 = 1 if there's any.
Solution. Applying the three conditions.
(I) 𝐺(1) = 3 + 𝑥 = 3 + 1 = 4
lim 𝐺(𝑥) = lim− (3 + 𝑥) = 3 + 1 = 4
𝑥→1− 𝑥→1
(II).
lim 𝐺(𝑥) = lim+ (3 − 𝑥) = 3 − 1 = 2
𝑥→1+ 𝑥→1
Because 𝐺 1 ≠ lim 𝐺 𝑥 (III) not satisfied.
𝑥→1
Hence, 𝐺 has discontinuity at 𝑥 = 1
𝑥 This type of discontinuity is called jump discontinuity.
Also an essential discontinuity.
Basic Calculus
PUPSHS
Continuity of
a Function Theorem 5. Theorem
If 𝑓 and 𝑔 are two functions continuous at 𝑥 = 𝑎, then
(I). 𝑓 + 𝑔 is continuous at 𝑎;
(II). 𝑓 − 𝑔 is continuous at 𝑎;
(III). 𝑓 ∙ 𝑔 is continuous at 𝑎;
(IV) 𝑓/𝑔 is continuous at 𝑎, provided that 𝑔(𝑎) ≠ 0.
Theorem 6. Theorem
A polynomial function is continuous at every
number.
Theorem 7. Theorem
A rational function is continuous at every number on its domain.
Basic Calculus
PUPSHS
Seatwork #2 Do the following.
1. Show that the absolute value function is continuous at 𝑥 = 0.
2. Show that the signum function is discontinuous at 𝑥 = 0.
3. Find the value of 𝑘 so that the function
𝑘𝑥 − 1 if 𝑥 ≤ 2
𝐺 𝑥 =ቊ
𝑘𝑥 2 if 𝑥 ≥ 2
is continuous everywhere.
Basic Calculus
PUPSHS
