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Limits & Continuity
Raymond LapusMathematics Department, De La Salle University
21 February 2012
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Contents
Limit of a function: Tabular & Graphical approach
Rules on evaluating limit of a function One-sided limits
Continuity of a function
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Motivation
ProblemA person earning PhP 150 per hour is paid only for the actual timeon the job. How close to 8 hours must a person work in order toearn within 3 pesos of the persons daily salary of PhP 1,200?
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Motivation
ProblemA person earning PhP 150 per hour is paid only for the actual timeon the job. How close to 8 hours must a person work in order toearn within 3 pesos of the persons daily salary of PhP 1,200?
Solution
x ... number of hours of work in a day
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Motivation
ProblemA person earning PhP 150 per hour is paid only for the actual timeon the job. How close to 8 hours must a person work in order toearn within 3 pesos of the persons daily salary of PhP 1,200?
Solution
x ... number of hours of work in a day f(x) = 150x ... amount paid per day in pesos at x hours of
work at the rate of PhP 150 per hour
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Motivation
ProblemA person earning PhP 150 per hour is paid only for the actual timeon the job. How close to 8 hours must a person work in order toearn within 3 pesos of the persons daily salary of PhP 1,200?
Solution
x ... number of hours of work in a day f(x) = 150x ... amount paid per day in pesos at x hours of
work at the rate of PhP 150 per hour
We solve for d > 0 such that |x 8| < d implies
|f(x)
1200
|< 3.
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Motivation
ProblemA person earning PhP 150 per hour is paid only for the actual timeon the job. How close to 8 hours must a person work in order toearn within 3 pesos of the persons daily salary of PhP 1,200?
Solution
x ... number of hours of work in a day f(x) = 150x ... amount paid per day in pesos at x hours of
work at the rate of PhP 150 per hour
We solve for d > 0 such that |x 8| < d implies
|f(x)
1200
|< 3.
|f(x) 1200| = |150x 1200| = |150(x 8)| = 150|x 8|.
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Motivation
ProblemA person earning PhP 150 per hour is paid only for the actual timeon the job. How close to 8 hours must a person work in order toearn within 3 pesos of the persons daily salary of PhP 1,200?
Solution
x ... number of hours of work in a day f(x) = 150x ... amount paid per day in pesos at x hours of
work at the rate of PhP 150 per hour
We solve for d > 0 such that |x 8| < d implies
|f(x)
1200
|< 3.
|f(x) 1200| = |150x 1200| = |150(x 8)| = 150|x 8|. 150|x 8| < 3 gives us |x 8| = 0.02.
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Motivation
ProblemA person earning PhP 150 per hour is paid only for the actual time
on the job. How close to 8 hours must a person work in order toearn within 3 pesos of the persons daily salary of PhP 1,200?
Solution
x ... number of hours of work in a day f(x) = 150x ... amount paid per day in pesos at x hours of
work at the rate of PhP 150 per hour
We solve for d > 0 such that |x 8| < d implies
|f(x)
1200
|< 3.
|f(x) 1200| = |150x 1200| = |150(x 8)| = 150|x 8|. 150|x 8| < 3 gives us |x 8| = 0.02. Hence, d = 0.02.
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Motivation
ProblemA person earning PhP 150 per hour is paid only for the actual time
on the job. How close to 8 hours must a person work in order toearn within 3 pesos of the persons daily salary of PhP 1,200?
Solution
x ... number of hours of work in a day f(x) = 150x ... amount paid per day in pesos at x hours of
work at the rate of PhP 150 per hour
We solve for d > 0 such that |x 8| < d implies
|f(x)
1200
|< 3.
|f(x) 1200| = |150x 1200| = |150(x 8)| = 150|x 8|. 150|x 8| < 3 gives us |x 8| = 0.02. Hence, d = 0.02. Therefore, a person must work within 1.2 minutes of 8
hours to earn within 3 pesos of his daily salary.
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Concept of the limit
The limit of a function is a means by which you can describe thebehaviour of a function as the independent variable x gets veryclose to a fixed number, say a.
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Concept of the limit
The limit of a function is a means by which you can describe thebehaviour of a function as the independent variable x gets veryclose to a fixed number, say a.
Example 1
The supply function for cooking oil is given by
p = g(x) =0.01x2 2500
0.1x 50where p is in pesos and x is the volume of cooking oil in 100 liters.Estimate the price per liter of cooking oil when there are close to50,000 liters supplied in the market.
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Limit of a function: Example 1 (contd)
Attention!
p = g(x) =0.01x2 2500
0.1x 50 = 0.1x + 50, x= 50.
Table of values (to approximate g(x) when x approaches 500)
x g(x)
499 99.9
499.5 99.95499.95 99.995
499.995 99.9995
499.9995 99.99995
x g(x)
501 100.1
500.5 100.05500.05 100.005
500.005 100.0005
500.0005 100.00005
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Limit of a function: Example 1 (contd)
Observation. As the value of x gets closer to 500 from eitherside, the corresponding function g(x) gets closer to 100.
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Limit of a function: Example 1 (contd)
Observation. As the value of x gets closer to 500 from eitherside, the corresponding function g(x) gets closer to 100. We writethis as
limx500
g(x) = limx500
0.01x2
25000.1x 50 = 100.
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Limit of a function: Example 1 (contd)
Observation. As the value of x gets closer to 500 from eitherside, the corresponding function g(x) gets closer to 100. We writethis as
limx500
g(x) = limx500
0.01x2
25000.1x 50 = 100.
Conclusion. (from the given problem)When there are close to 50,000 liters of cooking oil supplied in the
market, the price per liter of cooking oil gets closer to PhP 100.
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Limit of a function: Example 2
Non existence of a limitConstruct the table of values to determine whether the limit of the
functionf(x) =
x 20|x 20|
as x gets closer to 20 on either side exists or not.
Table of values (to approximate f(x) when x approaches 20)x f(x)
19 119.5 119.95 119.995 119.9995 1
x f(x)
21 1
20.5 1
20.05 120.005 1
20.0005 1
As x gets closer to the left of 20, f(x) gets closer to 1. On theother hand, as x gets closer to the right of 20, f(x) gets closer to1. Hence, the limit of f(x) does not exist.
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Exercise
Using a calculator, construct the table of values for x and f(x).
Use the values of x that approach the given number a from eitherside. Make a conclusion about lim
xaf(x) using this table.
1. limx1
f(x) where f(x) =
1 x2 if x= 1
2 if x = 1
2. limx1
f(x) where f(x) =
x2 if x < 10 if x 1
3. limx2
f(x) where f(x) =
4 x2 if 2 x < 2
x 2 if x 2
4. limx1
x2 1x2 + x
5. limx3
[[x 3]]
B i li i
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Basic limits
Basic limit formulaeFor all real numbers n = 0,
limxa
xn = an.
In particular,
limxa
x = a.
R l f l i li i
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Rules for evaluating limits
1. limit of a constant function
2. limit of a constant multiple of a function
3. limit of sum/difference of two functions
4. limit of product of two functions5. limit of quotient of two functions
6. limit of n-th power of a function
7. limit of n-th root of a function
8. limit of a polynomial function
LR1 Li it f t t f ti
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LR1: Limit of a constant function
If a and k are real numbers, then
limxa
k = k.
LR1 Li it f t t f ti
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LR1: Limit of a constant function
If a and k are real numbers, then
limxa
k = k.
Example 3 lim
x32w = 2w
limx0
5 = 5
limyrx = x
LR2 Li it of a co sta t lti le of a f ctio
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LR2: Limit of a constant multiple of a function
Let f be a function, a and k be real numbers. if limxa
f(x) = L, then
limxa
k f(x) = kL.
LR2: Limit of a constant multiple of a function
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LR2: Limit of a constant multiple of a function
Let f be a function, a and k be real numbers. if limxa
f(x) = L, then
limxa
k f(x) = kL.
Example 4
limx2
3x3 = 3
limx2
x3
= 3(8) = 24.
limy1
2y1/5 =
2 lim
y1
y1/5 = 2(1) = 2.
LR3: Limit of the sum/difference of functions
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LR3: Limit of the sum/difference of functions
Let f and g be functions and a be a constant. If limxa
f(x) = L and
limxa
g(x) = M, then
limxa
[f(x) g(x)] = L M.
LR3: Limit of the sum/difference of functions
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LR3: Limit of the sum/difference of functions
Let f and g be functions and a be a constant. If limxa
f(x) = L and
limxa
g(x) = M, then
limxa
[f(x) g(x)] = L M.
Example 5Let f(w) = 3w6, g(w) = 4w 2 and h(w) = 7. Evaluatelimw1
[f(w) + g(w) h(w)].
LR3: Limit of the sum/difference of functions
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LR3: Limit of the sum/difference of functions
Let f and g be functions and a be a constant. If limxa
f(x) = L and
limxa
g(x) = M, then
limxa
[f(x) g(x)] = L M.
Example 5Let f(w) = 3w6, g(w) = 4w 2 and h(w) = 7. Evaluatelimw1
[f(w) + g(w) h(w)].
Solutionlimw1
f(w) = 3,
LR3: Limit of the sum/difference of functions
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LR3: Limit of the sum/difference of functions
Let f and g be functions and a be a constant. If limxa
f(x) = L and
limxa
g(x) = M, then
limxa
[f(x) g(x)] = L M.
Example 5Let f(w) = 3w6, g(w) = 4w 2 and h(w) = 7. Evaluatelimw1
[f(w) + g(w) h(w)].
Solutionlimw1
f(w) = 3, limw1
g(w) = 2 and
LR3: Limit of the sum/difference of functions
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LR3: Limit of the sum/difference of functions
Let f and g be functions and a be a constant. If limxa
f(x) = L and
limxa
g(x) = M, then
limxa
[f(x) g(x)] = L M.
Example 5Let f(w) = 3w6, g(w) = 4w 2 and h(w) = 7. Evaluatelimw1
[f(w) + g(w) h(w)].
Solutionlimw1
f(w) = 3, limw1
g(w) = 2 and limw1
h(w) = 7.
Thus, limw1
[f(w) + g(w) h(w)] = 3 + 2 7 = 2.
LR4: Limit of the product of functions
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LR4: Limit of the product of functions
Let f and g be functions and a be a constant. If limxa
f(x) = L and
limxa
g(x) = M, then
limxa
[f(x) g(x)] = L M.
LR4: Limit of the product of functions
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LR4: Limit of the product of functions
Let f and g be functions and a be a constant. If limxa
f(x) = L and
limxa
g(x) = M, then
limxa
[f(x) g(x)] = L M.
Example 6
Let f(v) = 3v6 and g(v) = 2v + 9. Evaluate limv1
[f(v) g(v)].
LR4: Limit of the product of functions
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LR4: Limit of the product of functions
Let f and g be functions and a be a constant. If limxa
f(x) = L and
limxa
g(x) = M, then
limxa
[f(x) g(x)] = L M.
Example 6
Let f(v) = 3v6 and g(v) = 2v + 9. Evaluate limv1
[f(v) g(v)].
Solution
limv1 f(v) = 3 and
LR4: Limit of the product of functions
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p
Let f and g be functions and a be a constant. If limxa
f(x) = L and
limxa
g(x) = M, then
limxa
[f(x) g(x)] = L M.
Example 6
Let f(v) = 3v6 and g(v) = 2v + 9. Evaluate limv1
[f(v) g(v)].
Solution
limv1 f(v) = 3 and limv1 g(v) = 7.Thus, lim
v1[f(v) g(v)] = 3(7) = 21.
LR5: Limit of the quotient of two functions
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q
Let f and g be functions and a be a constant. If limxa
f(x) = L and
limxa
g(x) = M, then
limxa
f(x)
g(x)=
L
M, M= 0.
LR5: Limit of the quotient of two functions
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q
Let f and g be functions and a be a constant. If limxa
f(x) = L and
limxa
g(x) = M, then
limxa
f(x)
g(x)=
L
M, M= 0.
Example 7Let f(x) = 3 and g(x) = 2x 7. Evaluate lim
xu
f(x)
g(x)
where
u= 7/2.
LR5: Limit of the quotient of two functions
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q
Let f and g be functions and a be a constant. If limxa
f(x) = L and
limxa
g(x) = M, then
limxa
f(x)
g(x)=
L
M, M= 0.
Example 7Let f(x) = 3 and g(x) = 2x 7. Evaluate lim
xu
f(x)
g(x)
where
u= 7/2.
Solutionlimxu
f(x) = 3 and
LR5: Limit of the quotient of two functions
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Let f and g be functions and a be a constant. If limxa
f(x) = L and
limxa
g(x) = M, then
limxa
f(x)
g(x)=
L
M, M= 0.
Example 7Let f(x) = 3 and g(x) = 2x 7. Evaluate lim
xu
f(x)
g(x)
where
u= 7/2.
Solutionlimxu
f(x) = 3 and limxu
g(x) = 2u 7.
Thus, limxu
f(x)
g(x)
=
3
2u 7 .
LR6: Limit ofn-th power of a function
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Let f be functions, a be a real number and n be a positive integer.
If limxa
f(x) = L, then
limxa
(f(x))n = Ln.
LR6: Limit ofn-th power of a function
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Let f be functions, a be a real number and n be a positive integer.
If limxa
f(x) = L, then
limxa
(f(x))n = Ln.
Example 8Let f(y) = y3 12. Evaluate lim
y2[f(y)]3.
LR6: Limit ofn-th power of a function
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Let f be functions, a be a real number and n be a positive integer.
If limxa
f(x) = L, then
limxa
(f(x))n = Ln.
Example 8Let f(y) = y3 12. Evaluate lim
y2[f(y)]3.
Solution
limy2 f(y) = 2
3
12 = 4.Thus, lim
y2[f(y)]3 = (4)3 = 64.
LR7: Limit ofn-th root of a function
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Let f be functions, a be a real number and n be a positive integer
greater than or equal to 2. If limxa f(x) = L, then
limxa
n
f(x) =n
L.
LR7: Limit ofn-th root of a function
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Let f be functions, a be a real number and n be a positive integer
greater than or equal to 2. If limxa f(x) = L, then
limxa
n
f(x) =n
L.
Example 9Let f(y) = y3 11. Evaluate lim
y3
4
f(y).
LR7: Limit ofn-th root of a function
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Let f be functions, a be a real number and n be a positive integer
greater than or equal to 2. If limxa f(x) = L, then
limxa
n
f(x) =n
L.
Example 9Let f(y) = y3 11. Evaluate lim
y3
4
f(y).
Solution
limy3 f(y) = 3
3
11 = 16.Thus, lim
y3
4
f(y) =4
16 = 2.
LR9: Limit of a polynomial functionDi b i i f l i l
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Direct substitution property for polynomials.
If f is a polynomial function defined by
f(x) =
nj=1
bjxj,
then for any real number a,
limxa f(x) =
nj=1 b
ja
j
= f(a).
Example 10
Evaluate limx1
(x3 6x2 + 11x 6).
SolutionLet f(x) = x3 6x2 + 11x 6. We have
limx1
f(x) = f(1) = 13
6(12) + 11(1)
6 = 0.
Exercise
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Use the appropriate limit rules to evaluate each limit.
1. limx4
3x 22x + 5
2. limw1
(4w2 + 3w)4
3. limy5(2y2 8y + 3w)
4. limz3
(2z + 4)2(z + 3)4
(2 z)3
5. limx3 x2 + 5x + 6
3x2 + 14x + 15
6. limw2
w 1 1
w 27. lim
y1
y3 + y2 + 2y 4y2 3y + 2
8. limz5
3
z2 25
z3 125
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