Lecture on Limits

7/31/2019 Lecture on Limits

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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


Solution

x ... number of hours of work in a day
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Motivation


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


Solution



We solve for d > 0 such that |x 8| < d implies

|f(x)

1200

|< 3.
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Motivation


Solution




|f(x)

1200

|< 3.

|f(x) 1200| = |150x 1200| = |150(x 8)| = 150|x 8|.


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Motivation


Solution




|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




|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




|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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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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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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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]]

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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.

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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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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.

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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)].



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f(x) = L and

limxa

g(x) = M, then

limxa

[f(x) g(x)] = L M.


[f(w) + g(w) h(w)].

Solutionlimw1

f(w) = 3,

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f(x) = L and

limxa

g(x) = M, then

limxa

[f(x) g(x)] = L M.


[f(w) + g(w) h(w)].

Solutionlimw1

f(w) = 3, limw1

g(w) = 2 and



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f(x) = L and

limxa

g(x) = M, then

limxa

[f(x) g(x)] = L M.


[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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f(x) = L and

limxa

g(x) = M, then

limxa

[f(x) g(x)] = L M.

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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)].

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f(x) = L and

limxa

g(x) = M, then

limxa

[f(x) g(x)] = L M.

Example 6


[f(v) g(v)].

Solution

limv1 f(v) = 3 and



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p


f(x) = L and

limxa

g(x) = M, then

limxa

[f(x) g(x)] = L M.

Example 6


[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


f(x) = L and

limxa

g(x) = M, then

limxa

f(x)

g(x)=

L

M, M= 0.

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q


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.

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q


f(x) = L and

limxa

g(x) = M, then

limxa

f(x)

g(x)=

L

M, M= 0.


xu

f(x)

g(x)

where

u= 7/2.

Solutionlimxu

f(x) = 3 and

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f(x) = L and

limxa

g(x) = M, then

limxa

f(x)

g(x)=

L

M, M= 0.


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.

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If limxa

f(x) = L, then

limxa

(f(x))n = Ln.

Example 8Let f(y) = y3 12. Evaluate lim

y2[f(y)]3.



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If limxa

f(x) = L, then

limxa

(f(x))n = Ln.


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.

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limxa

n

f(x) =n

L.


y3

4

f(y).

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limxa

n

f(x) =n

L.


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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Lecture on Limits

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