(chapter 5) Limits

QQM 1023 Managerial Mathematics

lim f(x) xÆ a

6.1 INTRODUCTION TO LIMITS Perhaps you have heard about speed limit. What does the word “limit” means to you?? Or maybe you have been in a parking-lot situation in which you must “inch up” to the car in front, but yet you do not want to bump or touch it. This notation of getting closer and closer to something, but yet not touching it, is very important in mathematics and is involved in the concept of limits. Basically, we will let a variable “inch up” to a particular value and examine the effect it has on the values of a function.

6.2 DEFINITION OF LIMITS

Limits is based on the idea “getting closer to something” or “approaching something but yet not touching it”. •

The limit of f(x) as x approaches c is the number L, written

lim f ( x) = L

x →c

provided that f(x) is arbitrarily close to L for all x sufficiently close to, but not equal to, c.

Chapter 6: Limits

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QQM 1023 Managerial Mathematics •

We emphasize that, when finding a limit, we are concerned not with what happens to f(x) when x equals c, but only with what happens to f(x) when x is close to c.

•

Moreover, a limit must be independent of the way in which x approaches c. That is, the limit must be the same whether x approaches c from the left or from the right(for respectively).

•

x

< c or

x

> c,

For example: Let say; ƒ(x)

= 2x + 1, the limit for this function as x is

approaching the value of 1 is denoted by:

lim f ( x ) = lim(2 x + 1) x→2

•

x→2

In this case, x can approach 1 from both side: i)

Left : denoted by

x → 2− , means x can take values that are

less than and close to 2 (1.9, 1.99, 1.999, 1.9999….etc) ii)

Right: denoted by

x → 2+ , means that x can take values

that are greater and closer to 2 (2.1, 2.01, 2.001, 2.0001…)

•

There are three ways for us to obtain the limits of a function: ÆUsing the table of values ÆFrom the graph ÆUsing Algebra

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6.3

ESTIMATING LIMIT OF A FUNCTION: USING THE TABLE

Suppose that we are given a function,

f ( x) = 3x 2 − 1 and we would like to examine what is the value (limit) of this function as the input (x) approaches 4. In other word we try to solve this question;

(

)

lim f ( x ) = lim 3x 2 − 1 x →4

x →4

As we know, x can approach 4 in either way, left or right;

Step 1:

x approach 4 from left: x → 4 −

therefore the value of

x

must be less

than but close to 4:

x = 3.9 , x = 3.99 ,

x = 3.999 , x = 3.9999 Closer to 4

Closest to 4

Closer to 4 Less than

Substitute the selected values of

When

x = 3.9

When x

= 3.99

Chapter 6: Limits

x

to examine the effect on

f (x):

f ( x ) = 3x 2 − 1 f (3.9 ) = 3(3.9 )2 − 1 = 44.63 f ( x ) = 3x 2 − 1 f (3.99 ) = 3(3.99 )2 − 1 = 46.7603

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x = 3.999 and x = 3.9999 . What can you conclude?? Æ as x approach 4 (from left) f(x) will approach ??

Carry on with value of x that are closer to 4 like

You can also observe the limit for a function through table:

x f (x)

3.9

3.99

3.999

3.9999

x → 4− From the table, we can see that, as x approach 4 from the left, f ( x ) Table:

is getting closer (approach) to the value of 47 or

(

)

lim− f ( x ) = lim− 3x 2 − 1 ≈ 47

x→4

x→4

Step 2: Similarly, now let us examine if x approach 4 from the right:

x

where the values of

x → 4+

must be greater than 4. Select a few values of

x

that are greater than 4 (but remember it must be close to 4) like

x = 4.1, x = 4.01, x = 4.001 and x = 4.0001. Examine the effects on f(x) from the table:

x

4.1

4.01

4.001

4.0001

f (x) Table:

x → 4+

From the table, we can see that, as

x

approach 4 from the right,

is getting closer (approach) to the value of 47 or

(

f (x)

)

lim+ f ( x ) = lim+ 3x 2 − 1 ≈ 47

x→4

Chapter 6: Limits

x →4

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QQM 1023 Managerial Mathematics Step 3: As

lim f ( x )

=

x→4−

EQUAL

lim f ( x ) ≈ 47

x→4+

(

)

lim f ( x ) = lim 3x 2 − 1 = 47

Therefore we can conclude x → 4

x →4

DEFINITION 6.3.1: The limit of a function exist, if and only if the limit of the function measured from both side (left/right) are equal.

NOTES:

If

lim f ( x ) ≠ lim+ f (x ) ,

x→a −

x→a

therefore

lim f ( x ) DOES NOT EXIST x→ a

a = any value

Chapter 6: Limits

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Example 1:

x3 − 1 f (x) = , find the limit for f ( x ) as x x −1

Given

approach 1.

Solution :

Step 1: Build the table for x approaching 1 from the left :

x

0.9

0.99

x <1

0.999

0.9999

f (x) Table :

As

x

x → 1−

f (x )

approaching 1 from the left, we can see that

is

approaching 3 or:

 x3 −1  ≈ 3 lim f ( x ) = lim−  x →1− x →1 − 1 x   Step 2: Build the table for x approaching 1 from the right :

x

1.1

1.01

x >1

1.001

1.0001

f (x) Table :

As

x

x → 1+

approaching 1 from the right, we can see that

f (x )

is

approaching 3 or:

 x3 −1  ≈ 3 lim f ( x ) = lim+  x →1+ x →1  x −1  Chapter 6: Limits

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QQM 1023 Managerial Mathematics Step 3: As

Therefore,

lim f ( x )

x →1−

=

EQUAL

lim f ( x ) ≈ 3

x →1+

 x3 −1  = 3 lim f ( x ) = lim x →1 x →1 − 1 x  

Example 2: Determine whether

1 x →3 ( x − 3)

lim

exist or not?

Solution :

Step 1: Build the table for x approaching 3 from the left :

x f (x)

2.9

2.99

Table :

As

x

x<3

2.999

2.9999

x → 3−

approaching 3 from the left, we can see that

f (x )

is

approaching -10000 or:

lim− f ( x ) = lim−

x →3

x →3

1 = −10000 ≈ ?? (x − 3)

(where the closer x gets to 3, the smaller

f ( x ) will be Æ approaching -

∝)

Chapter 6: Limits

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Step 2: Build the table for x approaching 3 from the right :

x

3.1

3.01

x>3

3.001

3.0001

f (x) Table :

As

x

x → 3+

approaching 3 from the right, we can see that

f (x )

is

approaching 10000 or:

lim+ f ( x ) = lim+

x→3

(where

x→3

1 = 10000 ≈ ?? (x − 3)

f ( x ) is getting bigger as x approach 3 from right)

Step 3: As

lim f ( x ) ≠ lim+ f ( x )

x →3−

x →3

1 = x →3 ( x − 3)

lim f ( x ) = lim

Therefore, x →3

Chapter 6: Limits

????

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6.4

ESTIMATING LIMIT OF A FUNCTION: FROM THE GRAPH

Another way to obtain the limit of a function is from the graph.

Example 3: Based on the graph for the function

f ( x ) , we can examine the

limit for this function as x approach 1:

y 6 5 4 3 2 1

-4

a)

-3

-2

-1

0

1

2

3

x

→ → x → 1−

1+ ← x ← ←

lim f (x )

ii)

Find i)

iii)

x →1−

lim f ( x )

x →1+

lim f ( x ) x →1

Chapter 6: Limits

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Example 4: y 6 5 4 3 2 1

-4

-3

-2

-1

0

1

→ → x → 2−

b)

2

3

x

2+ ← x ← ←

Find i)

iii)

lim f ( x )

x →2 −

ii)

lim f ( x )

x →2 +

lim f ( x ) x→2

Chapter 6: Limits

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Example 5: y 6 5 4 3 2 1

-4

-3

-2

→ → x → −1−

c)

-1

0

1

2

3

x

− 1+ ← x ← ←

Find i)

iii)

lim f ( x )

ii)

lim f (x )

iv)

( )

x → −1−

x →( −1)

Chapter 6: Limits

lim f ( x )

( )

x → −1+

f (− 1)

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QQM 1023 Managerial Mathematics

6.5

ESTIMATING LIMIT OF A FUNCTION: BY USING ALGEBRA

PROPERTIES OF LIMITS: PROPERTY 1

lim c = c

i) x → a

where c is a constant

Example 6: a)

lim 7 =

b)

lim e =

x→2

x →5

PROPERTY 2

lim x n = a n

ii) x → a

Example 7:

lim x 2 =

a) x →3

b)

c)

lim x1 / 3 =

x → 27

lim x1.3 = x →e

Chapter 6: Limits

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QQM 1023 Managerial Mathematics Let say:

lim f ( x ) = L

and

x→a

lim g ( x ) = M x→a

where L and M are real numbers, therefore: PROPERTY 3

lim[ f (x ) ± g ( x )] = lim f ( x ) ± lim g ( x )

iii) x → a

x→a

x→a

= L ± M PROPERTY 4

lim[ f ( x ).g (x )] = lim f (x ). lim g ( x )

iv) x → a

x→a

x→a

=LxM PROPERTY 5

lim[c. f (x )] = c. lim f (x )

v) x → a

x→a

=cL PROPERTY 6

lim n f ( x ) = n lim f ( x )

vi) x → a

x→a

=

Chapter 6: Limits

n

L

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QQM 1023 Managerial Mathematics

Example 8:

f ( x) = 2 x + 3

Given

and

a)

lim(2 x + 3) =

b)

lim( x 2 + 1) =

g ( x) = x 2 + 1 . Find

x→2

x→2

[

]

c)

lim (2 x + 3) − ( x 2 + 1) =

d)

lim(2 x + 3)( x 2 + 1) =

x→2

x→2

e)

lim e( x 2 + 1) =

f)

lim (2 x + 3) =

x→2

x→2

Chapter 6: Limits

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QQM 1023 Managerial Mathematics PROPERTIES OF LIMITS FOR RATIONAL FUNCTION: If:

lim f ( x ) = L

and

x →a

lim g ( x ) = M x→a

where L and M are real numbers, therefore:

The limit for the rational function :

f (x ) h(x) = g (x )

, is obtain by

f (x ) L lim h( x) = lim = x→a x → a g (x ) M However, there are 4 possibilities for the values of L and M that will define the limit for this function:

L ≠ 0 and M ≠ 0

POSSIBILITY 1

Therefore;

lim x→a

f (x ) L = ≠0 g (x ) M

Example 9: Given:

f ( x) = 5 x − 4

Find:

and

a) lim (5 x − 4)

=

b) lim(3 x + 7)

=

x →3

x →3

c)

g ( x) = 3x + 7

(5 x − 4) = x →3 (3 x + 7)

lim

Chapter 6: Limits

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QQM 1023 Managerial Mathematics

Example 10: a) lim x →3

b)

x−2 x −8

Find:

x2 − 5 = lim x →2 1 − x

L =0

POSSIBILITY 2

and M ≠ 0

Therefore;

lim x→a

f (x ) 0 = =0 g (x ) M

Example 11:

lim

a) x→4

b)

x−4 = x+2

2x − 4 = x →2 3

lim

L ≠ 0 and M = 0

POSSIBILITY 3

Therefore;

lim x→a

Chapter 6: Limits

f (x ) L = → g (x ) 0

DOES NOT EXIST

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QQM 1023 Managerial Mathematics

Example 12: a) lim x →3

b)

x = x−3

4x = x→3 2 x − 6

lim

POSSIBILITY 3

lim x→a

•

f (x ) 0 = → g (x ) 0

L = 0 and M = 0 LIMIT MIGHT/MIGHT NOT EXIST

To determine whether the limit exist/not, check whether we can simplify the function.

Example 13:

lim

a) x→2

b)

x−2 = x2 − 4

3x 2 − 7 x + 2 = lim x →2 x−4

Chapter 6: Limits

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QQM 1023 Managerial Mathematics CONCEPTS OF INFINITY (∞) i)

a+∞=∞

ii)

∞+a=∞

iii)

a - ∞ = -∞

iv)

∞-a=∞

v)

a.∞ = ∞

vi)

a(-∞) = - ∞

vii)

a=0 ∞

viii)

∞=∞ a

ix)

∞n = ∞; for n>0

x)

∞n = 0; for n<0

xi)

∞n = 1; for n=0

xii)

n

∞

=∞

* a is constant 6.5.1 LIMIT AT INFINITY (∞) i) Limits at infinity for a constant function:

lim a = a x →∞

where a is constant.

Limit of a constant function will always remain no matter what value does x approaching.

Example 14: Find:

π a) lim x →∞ b)

=

lim 2.718 =

x → −∞

Chapter 6: Limits

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QQM 1023 Managerial Mathematics ii) Limit at infinity for polynomial function:

lim ax n = a.∞ n x →∞

where a is constant.

To find limit at infinity for a polynomial function, simply substitutes the input (x) with ∝.

Example 15:

(3 x a) lim x →∞

b)

2

Find;

+ 1) =

lim(2 x 5 + x 4 − 1000) = x→∞

iii) Limit at infinity for rational function: h( x) =

f ( x) → numerator g ( x) → deno min ator

Steps to obtain limit at infinity for rational function; i. Exclude all terms except the one with the greatest power for numerator and denominator. ii.

Simplify the remaining terms

iii.

Substitutes x with ∞ or -∝

iv.

Find the limit for the function.

Chapter 6: Limits

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Example 16: a)

2x4 + x2 − 3 = lim 3 x→∞ x + x + 2

b)

2 x3 + x 2 − 3 = lim 3 x→∞ x + x + 2

x +1 = + 5x

c) lim x→∞ 2 x 2

Chapter 6: Limits

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