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Maths-Module 3 Differential calculus
computer science (csc 104)
University of Embu
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University of Embu Department of Mathematics, Computing and Information Technology (MCIT) Course code: SMA 103 Course Title : Differential Calculus Semester: First Academic year: 2020/ Lecturer : Dr Cyrus Ngari MODULE 3
Infinite Limits
a. We say that the 𝑥→𝑎lim𝑓(𝑥) = ∞, if we can make 𝑓(𝑥) arbitrarily large for all values of 𝑥
sufficiently close to 𝑥 = 𝑎 from both sides without actually letting 𝑥 = 𝑎. b. Similarly, We say that the 𝑥→𝑎lim𝑓(𝑥)= −∞, if we can make 𝑓(𝑥) arbitrarily large and
negative for all values of 𝑥 sufficiently close to 𝑥 = 𝑎 from both sides without actually letting 𝑥 = 𝑎.
Example 4:
- Evaluate lim𝑥→0 1 𝑥 Solution: here direct substitution cannot be done to avoid the denominator becoming zero. To proceed, we need to evaluate the limit from the left and from the right and observe the behavior for conclusion;
𝑥 −0. 01 −0. 001 −0. 0001 −0. 00001 0 0. 00001 0. 0001 0. 001 0. 01 𝐹(𝑥)
=
1
𝑥
−100 −1000 −10,000 −100,0000? 100,000 10,000 1000 100
Clearly, the table shows that:
Row one shows that 𝑥 approaches 0 from the right (above) and from the left (below) respectively. Row two shows that 𝑓(𝑥) approaches +∞ and −∞ from the right (above) and from the left (below) respectively. Hence lim𝑥→ 0 −
1 𝑥= −∞ and 𝑥→lim 0 +
1 𝑥= +∞ Now since 𝑥→lim 0 −
1 𝑥= −∞ ≠ 𝑥→lim 0 +
1 𝑥= +∞ (limit from the left(below) is not equal to the
limit from the right (above)), we say that the limit of the function 1 𝑥 as 𝑥 approaches 0 does not exist.
The graph of the function 1 𝑥 as 𝑥 approaches 0 behaves as the below graph show.
Exercise 4: Evaluate the following limits
a. lim𝑥→0𝑥 62
b. 𝑥→−2lim𝑥+2−
c. lim𝑥→4(4−𝑥 3 ) 3
Definition: A function 𝑓(𝑥) is said to be continuous at point 𝑥 = 𝑐 if the following 3 conditions
are satisfied:
i. 𝑓(𝑐) must be defined i 𝑐 must be in the domain of 𝑓(𝑥). ii. The lim𝑥→𝑐𝑓(𝑥) must exist. iii. The lim𝑥→𝑐𝑓(𝑥) = 𝑓(𝑐).
Note: If one of the above conditions is not satisfied by any function, then the particular function
is said to be discontinuous.
Remarks:
a. A function 𝑓(𝑥) is continuous on the open interval (𝑎,𝑏) if it is continuous at each point in the interval. b. A function that is continuous on the entire real line (−∞,∞) is everywhere continuous. c. Geometrically, the graph of 𝑦 = 𝑓(𝑥) drawn for a given interval should not have jumps, breaks or holes.
Example 4: Determine whether the function 𝑓(𝑥)= 𝑥 2 at 𝑥 = 1 is continuous.
Solution: We need to show that the function satisfies the three conditions for continuity. At 𝑥 =
1 , it implies that 𝑐 = 1, thus;
a. 𝑓(𝑐)= 𝑓( 1 )= 1 2 = 1, hence 𝑓(1) is defined.
b. The lim𝑥→1𝑥 2 =1 2 = 1, hence the limit exist.
c. lim𝑥→1𝑥 2 = 𝑓( 1 )= 1, hence satisfied.
Therefore, we conclude that 𝑓(𝑥)= 𝑥 2 is continuous at 𝑥 = 1since it satisfies all the three
conditions for continuity.
Example 4: Test the continuity of 𝑓(𝑥)= 1 𝑥 at 𝑥 = 0.
Solution: We test whether the function satisfies the conditions for continuity
a. 𝑥→lim 0 −
1 𝑥= −∞ and 𝑥→lim 0 +
1 𝑥= +∞, clearly, the limit does not exist at 𝑥 = 0 since 𝑥→lim 0 −
1 𝑥= −∞
≠ 𝑥→lim 0 +
1 𝑥= +∞.
Therefore, without even testing for the satisfaction of the other two conditions, we conclude
immediately that the function is discontinuous.
Example 4: Determine the continuity of the Heaviside function defined by;
𝐻(𝑡)= {0 𝑖𝑓 𝑡 < 01 𝑖𝑓 𝑡 ≥ 0}
Solution: Checking the satisfaction of the conditions for continuity we have,
Clearly,
lim𝑡→ 0 −
𝐻(𝑡) = 0 and lim𝑡→ 0 +
𝐻(𝑡) = 1
Now since
lim𝑡→ 0 −
𝐻(𝑡) = 0 ≠lim𝑡→ 0 +
𝐻(𝑡) = 1 it implies that the limit does not exist. Thus the function is not
continuous.
Type of Discontinuity
The following are two types of discontinuity:
ii. 𝑓(𝑥)= 𝑥 2 − 6𝑥 + 9 at 𝑥 = 3.
iii. 𝑓(𝑥)=𝑥
2 − 𝑥+4 at 𝑥 = −4. 2. Discuss the continuity of the following hence state the type of discontinuity.
a. 𝑥→−3lim𝑥
2 − 𝑥+ b. lim𝑥→1𝑥
2 − 𝑥− c. 𝑥→−2lim𝑥
2 +4𝑥+ 𝑥+
d. 𝐻(𝑡)= {0 𝑖𝑓 𝑡 < 01 𝑖𝑓 𝑡 ≥ 0}
e. 𝑓(𝑥)= {
𝑥 𝑖𝑓 0 ≤ 𝑥 < 1
1 2 𝑥 𝑖𝑓 1 ≤ 𝑥 < 0
}
f. 𝑓(𝑥)= {1 − 𝑥 𝑥 𝑖𝑓 𝑥 > 1𝑖𝑓 𝑥 ≤ 1}
g. 𝑓(𝑥)=𝑥 12 at 𝑥 = 0. h. 𝑓(𝑥)= sin 𝑥
i. 𝑓(𝑥)=𝑥
2 − 𝑥+9.
Limits of Trigonometric Functions
Theorem 4: Squeeze Theorem: This theorem states that; lim𝑥→0sin𝑥𝑥 = 1.
Clearly, from the Squeeze theorem we can deduce that 𝑥→0limsin𝑥𝑥 = [lim𝑥→0sin𝑥𝑥 ]
− = 1. The theorem
is applied in solving problems involving limits of trigonometric functions.
Examples 4: Evaluate the following
a. lim𝑥→0sin𝑥5𝑥 Solution:
𝑥→0lim
sin𝑥 5𝑥 = lim𝑥→0[
sin𝑥 5𝑥 ×
5
5 ] = lim𝑥→
sin5𝑥 5𝑥 × lim𝑥→
1
5 = 1 ×
1
5 =
1
5.
b. lim𝑥→0sin
2 𝑥 𝑥 Solution:
lim𝑥→
sin 2 𝑥 𝑥 = lim𝑥→0[
sin𝑥 ∙ sin 𝑥 𝑥 ] = lim𝑥→
sin𝑥 𝑥 × lim𝑥→0sin𝑥 = 1 × 0 = 0. c. lim𝑥→0sin2𝑥5𝑥 Solution:
𝑥→0lim
sin2𝑥 5𝑥 = lim𝑥→0[sin2𝑥 ×
1
sin5𝑥×
2𝑥
2𝑥×
5
5 ] = lim𝑥→0[
sin2𝑥 2𝑥 ×
5𝑥
sin 5𝑥] ×
2
5
= lim𝑥→
sin2𝑥 2𝑥 × [lim𝑥→
5𝑥
sin5𝑥]
− = 1 × 1 ×
2
5 =
2
5.
d. Show that, 𝑥→0lim1−cos𝑥𝑥 = 0
Solution: Here we need to rationalize the numerator first in order to apply the squeeze theorem.
lim𝑥→
1 − cos 𝑥 𝑥 = lim𝑥→0[
1 − cos 𝑥 𝑥 ×
1 + cos 𝑥 1 + cos 𝑥]
= lim𝑥→
1 − cos 2 𝑥 𝑥(1 +cos 𝑥)
= lim𝑥→
sin 2 𝑥 𝑥(1 +cos 𝑥)
= lim𝑥→0[
sin𝑥 ∙ sin 𝑥 𝑥(1 +cos𝑥)] = lim𝑥→
sin𝑥 𝑥 × lim𝑥→
sin𝑥 1 + cos 𝑥= 1 × 0 = 0. Hence shown.
Exercise 4: Evaluate the following
lim𝑥→0sin𝑥3𝑥
lim𝑥→0sin5𝑥𝑥
lim𝑥→0sin2𝑥3𝑥
lim𝑥→0sin𝑥3𝑥
lim𝑥→0sin
2 𝑥 𝑥 2 6. lim𝑥→03(1−𝑥cos𝑥)
Maths-Module 3 Differential calculus
Course: computer science (csc 104)
University: University of Embu
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