Transcript Document

Algebra of Limits
Assume that both of the following limits exist and c and
is a real number:
Then:
1. . lim cf ( x)  c lim f ( x)
xa
xa
2. .lim f ( x)  g ( x)  lim f ( x)  lim g ( x)
xa
xa
xa
3. . lim f ( x) g ( x)  lim f ( x)  lim g ( x)
xa
xa
xa
4. . lim f ( x) / g ( x)  lim f ( x) / lim g ( x),
xa
xa
provided that lim g ( x)  0
xa
xa
Calculating Limits
Finding the limit of a function f a point x = a.
Distinguishing the following cases:
1. The case when f is continuous a x = a.
2. The case 0/0.
3. The case ∞/ ∞
4. The case of an infinite limit
5. The case c/∞, where c is a real number.
6. The case, when it is possible to use the
squeeze theorem.
1. The case when f is continuous at x = a
If f is continues at x=a, then:
Notice:
1. Polynomial functions and the cubic root function
( & all functions of its two families) are everywhere
continuous.
2. Rational, trigonometric and root functions are
continuous at every point of their domains.
3. If f and g are continuous a x=a, then so are cf, f+g,
f-g, fg and f/g (provided that he limit of f at x=a is
not zero)
Examples for the case when f is
continuous at x = a
Example (1)
x 4
lim
x  2 x  2
2
x 4
The rational function f ( x) 
is cont.
x2
on R  {2}, and so it is cont. at x  2
2
(2)  4
0
lim f ( x)  f (2) 

0
x  2
22
4
2
Examples for the case when f is
continuous at x = a
Example (2)



lim ( x 5  2 x  3) 9 x  9  x  3
x 1

The polynomial function p ( x)  x 5  2 x  3 is cont everywhere, thus it is cont. at x  1
The function s ( x)  9 x  9 is cont everywhere, thus it is cont. at x  1


Therefore the product function g ( x)  ( x 5  2 x  3) 9 x  9 is cont.at x  1
The root function h( x)  x  3 is cont on (3, ), thus it is cont. at x  1
Thus,




The function f ( x)  ( x 5  2 x  3) 9 x  9  x  3 is cont at x  1.
and so,
lim f ( x)  f (1)  ((1) 5  2(1)  3) 9 1  9  1  3
x 1
 (1  2  3)9  8  4  2(2)  2  2
Examples for the case when f is continuous at x= a
Example (3) :
lim3 x  x  5 
x 5
g ( x)  3 x and h( x)  x  5 are cont. everywhere
(Questions :
1. Show that h is cont. at x  5
2. Graph h )
and so they are cont.at x  5, thus :
f ( x)  3 x  x  5 is cont. at x  5
Thus
lim3 x  x  5   f (5)  15  5  5  15
x 5
2. The case 0/0
Suppose we want to find:
g ( x)
lim
x a h( x )
For the case when: lim g ( x) & lim h( x) are 0.
xa
xa
Then this is called the case 0/0. Caution:
The limit is not equal 0/0. This is just a
name that classifies the type of limits having
such property.
Examples for the case 0/0
Example (1) : Solving by factoring
x3  8
lim 4
x  2 x  16
This is the case 0 / 0, because :



lim x 3  8  0  lim x 4  16
x2
x2

x3  8
lim 4
x  2 x  16
( x  2)( x 2  2 x  4)
 lim
x  2 ( x  2)( x  2)( x 2  4)
x2  2x  4
4  4  4 12 3
 lim



2
x  2 ( x  2)( x  4)
4(8)
32 8
Examples for the case 0/0
Example (2) Multiplying by the conjugate method
x  25  5
lim
x 0
x( x  3)
This is the case 0 / 0, because :


lim x  25  5  0  lim x( x  3) 
x 0
x  25  5
lim
 lim
x 0
x 0
x( x  3)

x 0




x  25  5
.
x( x  3)
x  25  5
x  25  5
( x  25)  25
x
 lim
 lim
x 0 x ( x  3)
x  25  5 x 0 x( x  3) x  25  5
1
1
1
 lim


x 0 ( x  3)
x  25  5 3(5  5) 30






Examples for the case 0/0
Example (3) Involving Absolute Values
Findif exists : lim
x  5
2 x  10
3 x  15
Solution :
This is the case 0 / 0, because :
lim 2 x  10  0  lim 3 x  15
x  5
x  5
Notice that
2 x  10 
lim 
x ( 5 )
lim 
x ( 5 )

2 x 10
; x  5
 ( 2 x 10 )
; x  5
2 x  10
3 x  15
2 x  10
3 x  15
Thus : lim 
 lim 
2 x  10
2( x  5)
2 2
 lim 
 lim  
3 x  15 x ( 5) 3( x  5) x ( 5) 3 3
 lim 
 (2 x  10)
 2( x  5)
2
2
 lim 
 lim 

x ( 5 )
x ( 5 )
3 x  15
3( x  5)
3
3
x ( 5 )
x ( 5 )
2 x  10
 lim 
2 x  10
3 x  15 x ( 5) 3 x  15
2 x  10
 lim
Does not exist
x  5 3 x  15
x ( 5 )
Question: Simplify the formula of f and graph it!

f ( x) 

3 x  15 
2 x  10
2 x 10
3 x 15
 ( 2 x 10 )
3 x 15


x  5

; x  5
;
2
3
2
3
; x  5
; x  5
Questions
1.Let f 
x
x
a. Find , if either exists, the left limit , the left limit
and the limit at the po int x  0.
b. Simplify the formula of f
c. Graph f
2.Let f 
2x  6
x3
a. Find , if either exists, the left limit , the left limit
and the limit at the po int x  0.
b. Simplify the formula of f
c. Graph f
3. The case ∞/ ∞
Suppose we want to find:
g ( x)
g ( x)
lim
Or lim
x  a h( x )
x   h( x )
For the case when the limits of both functions f and g
are infinite
Then this is called the case ∞/ ∞. Caution:
The limit is not equal ∞/∞. This is just a
name that classifies the type of limits having
such property.
Limits at infinity
A function y=f(x) may approach a real number b
as x increases or decreases with no bound.
When this happens, we say that f has a limit at
infinity, and that the line y=b is a horizontal
asymptote for f.
Limit at infinity: The Case of Rational
Functions
A rational function r(x) = p(x)/q(x) has a limit at infinity if the
degree of p(x) is equal or less than the degree of q(x).
A rational function r(x) = p(x)/q(x) does not have a limit at
infinity (but has rather infinite right and left limits) if the degree
of p(x) is greater than the degree of q(x).
Example (1)
Let
5x9  2 x 2  3
f ( x)  9
6x  4x7  1
Find
lim f ( x)
x  
Solution:
Since the degree of the polynomial in the
numerator, which is 9, is equal to the degree of
the polynomial in the denominator, then
The cofficient of x 9 in the numerator
lim f ( x) 
x  
The cofficient of x 9 in the denominator
5

6
To show that, we follow the following steps:
1
3

7
9
5x9  2 x 2  3
x
x
lim f ( x)  lim 9
 lim
7
x  
x   6 x  4 x  1
x  
1
1
64 2  9
x
x
1
3

lim  5  2 7  9  lim 5  2 lim 1  3 lim 1
x  
x
x  x 
x   x 7
x   x 9



1
1
1
1 

lim
6

4
lim

lim
lim  6  4 2  9 
x  
x   x 2
x   x 9
x  
x
x


5  2(0)  3(0) 5


6  4(0)  0
6
52
Example (2)
Let
5x9  2 x 2  3
f ( x)  12
6x  4x7  1
Find
lim f ( x)
x  
Solution:
Since the degree of the polynomial in the numerator,
which is 9, is less than the degree of the polynomial
in the denominator, which is 12, then
lim f ( x)  0
x  
To show that, we follow the following steps:
5
1
3
 2 10  12
3
5x9  2 x 2  3
x
x
x
lim f ( x)  lim 12

lim
x  
x   6 x  4 x 7  1
x  
1
1
6  4 5  12
x
x
1
3 
 5
lim  3  2 10  12  5 lim 1  2 lim 1  3 lim 1
x   x
x
x 
x   x 3
x   x10
x   x12



1
1
1
1 

lim 6  4 lim 5  lim 12
lim  6  4 5  12 
x
 
x   x
x   x
x  
x
x 

5(0)  2(0)  3(0) 0

 0
6  4(0)  0
6
Example (3)
Let
5 x12  2 x 2  3
f ( x) 
6 x9  4 x7  1
Find
lim f ( x)
x  
Solution:
Since the degree of the polynomial in the numerator,
which is 12, is greater than the degree of the
polynomial in the denominator, which is 9, then
lim f ( x) do not exist .
x  
They are infinite limits. To show that, we follow the following steps:
5x  2 x  3
lim f ( x)  lim
x  
x   6 x 9  4 x 7  1
12
5x
are the same as lim
x   6 x 9
which are the same as lim x 3  
12
2
x  
Example (4)
Let
5 x12  2 x 2  3
f ( x) 
 6 x9  4 x7  1
Find
lim f ( x)
x  
Solution:
Since the degree of the polynomial in the numerator,
which is 12, is greater than the degree of the
polynomial in the denominator, which is 9, then
lim f ( x) do not exist .
x  
They are infinite limits. To show that, we follow the following steps:
5x  2 x  3
lim f ( x)  lim
x  
x    6 x 9  4 x 7  1
12
5x
are the same as lim
x    6 x 9
which are the same as lim ( x 3 )   
12
2
x  
Example (5)
Let
 5 x12  2 x 2  3
f ( x) 
 6 x8  4 x 7  1
Find
lim f ( x)
x  
Solution:
Since the degree of the polynomial in the numerator,
which is 12, is greater than the degree of the
polynomial in the denominator, which is 8, then
lim f ( x) do not exist.
x  
They are infinite limits. To show that, we follow the following steps:
 5x  2 x  3
lim f ( x)  lim
x  
x    6 x 8  4 x 7  1
12
 5x
are the same as lim
x    6 x 8
which are the same as lim x 4  
12
2
x  
Example (6)
Let
5 x12  2 x 2  3
f ( x) 
 6 x8  4 x 7  1
Find
lim f ( x)
x  
Solution:
Since the degree of the polynomial in the numerator,
which is 12, is greater than the degree of the
polynomial in the denominator, which is 8, then
lim f ( x) do not exist.
x  
They are infinite limits. To show that, we follow the following steps:
5x  2 x  3
lim f ( x)  lim
x  
x    6 x 8  4 x 7  1
12
5x
are the same as lim
x    6 x 8
which are the same as lim ( x 4 )  
12
2
x  
Limis & Infinity
Problems Involving Roots
Introduction
We know that:
√x2 = |x|
, which is equal x is x non-negative and
equal to – x if x is negative
For if x = 2, then √(2)2 = √4 = |2|=2
& if x = - 2, then √(-2)2 = √4 = |-2|=-(-2) = 2
Example
Let :
f ( x)  16 x 2  9
can be rewwritten as :
f ( x)  16 x 2  9
 x 2 (16 x92 )

3
x

3
(16 
9 )
x2
 x 16 9

2
   x 16x 9
x2
 3
;
x  0 or x  0
;
x 0
;
x  0 or x  0
;
x 0
;
x0
;
x 0
;
x 0
Example
Let :
16 x 2  9
2x  2
Find the horizontal asymptotes of f
f ( x) 
Solution :
We frst find the li mits of f at   and at  
We will find these lim its ogother in one formula,
lim f ( x)  lim
x  
x  
x
 lim
x  
16 
16 x  9
 lim
x  
2x  2
2
9
x2
2x  2
 x 16 
 lim
x  
x 2 (16 
9
)
x2
2x  2
9
x2
2
x(2  )
x
x2
 lim
 lim
x  
2x  2
x  
 16 
2
16 
9
x2
2
x
9
x  
x   x 2
 16  0  4



2
20
2
lim 2  lim
x  
x   x
 2 Why ?

lim 16  lim
Thus the lines y  2 and y  2 are the horizontal asymptotes of f
9
x2
4. The case of infinite limit
Infinite Limits
A function f may increases or decreases with no
bound near certain values c for the
independent variable x.
When this happens, we say that f has an infinite
limit, and that f has a vertical asymptote at x = c
The line x=c is called a vertical asymptote for f.
Infinite Limits- The Case of Rational
Functions
A rational function has an infinite limit at a
point x=c if at that point the limit of the
denominator is zero and the limit of the
numerator is not zero. The sign of the infinite
limit is determined by the sign of both the
numerator and the denominator at values close
to the considered point x=c approached by the
variable x.
Example (1)
Let
1
f ( x) 
x
Find
a. lim f ( x)
x 0
b. lim f ( x)
x 0
Solution:
First x=0 is a zero of the denominator which is
not a zero of the numerator.
a. As x approaches 0 from the right, the numerator is always positive ( it is equal to 1)
and the denominator approaches 0 while keeping positive; hence, the function
increases with no bound. Thus:
lim f ( x)  
x 0
The function has a vertical asymptote at x = 0, which is the line x = 0 (see the graph in
the file on basic algebraic functions).
b. As x approaches 0 from the left, the numerator is always positive ( it is equal to 1) and
the denominator approaches 0 while keeping negative; hence, the function decreases
with no bound.
lim f ( x)  
x 0
Example (2)
x5
f ( x) 
x 1
Let
Find
a. lim f ( x)
x 1
b. lim f ( x)
x 1
Solution:
First x=1 is a zero of the denominator which is
not a zero of the numerator.
a. As x approaches 1 from the right, the numerator approaches 6 (thus keeping positive),
and the denominator approaches 0 while keeping positive; hence, the function increases
with no bound. Thus,
lim f ( x)  
x 1
The function has a vertical asymptote at x = 1, which is the line x = 1
b. As x approaches 1 from the left, the numerator approaches 6 (thus keeping positive),
and the denominator approaches 0 while keeping negative; hence, the function decreases
with no bound. The function has a vertical asymptote at x=1, which is the line x = 1. Thus:
lim f ( x)  
x 1
Example (3)
Let
x 2  5 x  4 ( x  1)( x  4)
f ( x)  2

x  4 x  3 ( x  1)( x  3)
Find
a. lim f ( x)
x 3
b. lim f ( x)
x 3
Solution:
First x=3 is a zero of the denominator which is
not a zero of the numerator.
lim f ( x)
x 3 
x 2  5x  4
 lim 2
x 3 x  4 x  3
( x  1)( x  4)
 lim
x 3 ( x  1)( x  3)
x4
 lim
x 3 x  3
As x  3 x  4  3  4  1 ; thus keeping negative
and x  3  0 , while keeping positive
(because as x  3 , we have x  3 and so x  3  0 )
Thus lim f ( x)  
x 3
lim f ( x)
x 3
x 2  5x  4
 lim 2
x 3 x  4 x  3
( x  1)( x  4)
 lim
x 3 ( x  1)( x  3)
x4
 lim
x 3 x  3
As x  3 x  4  3  4  1 ; thus keeping negative
and x  3  0 , while keeping negative
(because as x  3 , we have x  3 and so x  3  0 )
Thus lim f ( x)  
x 3
6. The case constant/∞
Suppose we want to find:
g ( x)
lim
x a h( x )
For the case when:
lim g ( x)  c  R & lim h( x)  
xa
xa
In this case, no mater what the formulas of g and h are, we will always
have:
g ( x)
lim
0
x  a h( x )
Then this is called the case c/∞. Caution: The limit is not equal c/ ∞. This
is just a name that classifies the type of limits having such property. This
limit is always equal zero
Example on the case constant/∞
Example(1) :
Find ,
1
lim
x  
x 1  x
Solution :


We have :
g ( x)  1  lim g ( x)  lim 1  1
x  
x  
h( x)  x  1  x  lim h( x)   (Why ?)
x  
Thus,
g ( x)
lim
 lim
x   h( x )
x  

1
0
x 1  x

6. Using the Squeeze Theorem
The Squeeze (Sandwich or Pinching))
Theorem
Suppose that we want to find the limit of a
function f at a given point x=a and that the
values of f on some interval containing this
point (with the possible exception of that point) lie between the
values of a couple of functions g and h whose
limits at x=a are equal. The squeeze theorem
says that in this case the limit of f at x=a will
equal the limit of g and h at this point.
The Squeeze Theory
Let :
g ( x )  f ( x )  h( x )
; x  (c, d )  a
where a  (c, d )
&
lim g ( x)  l  lim h( x)
xa
Then :
lim f ( x)  l
xa
xa
Example (1)
Let :
2 x  1  f ( x)  x 2
Find
; x  (0,4)
lim f ( x)
x 1
Soluion :
We have :
lim (2 x  1)  2  1  1 &
x 1
2
2
x

(
1
)
1
lim
x 1
&
1  (0,4) , 2 x  1  f ( x)  x 2 ; x  (0,4)
Thus, by the squeeze theorem,
lim f ( x)  1
x 1
Example (2)
Let :
4 x  9  f ( x)  x 2  4 x  7 ; x  [0, )
Find
lim f ( x)
x4
Soluion :
We have :
lim (4 x  9)  16  9  7
,
x4
2
(
x
lim  4 x  7)  16  16  7  7
x4
&
4  (0, ) , 2 x  1  f ( x)  x 2
Thus, by the squeeze theorem,
lim f ( x)  7
x4
; x  [0, )
Example (3)
Find
1
2
(
x
sin
)
lim
x
x 0
Soluion :
:
We have :
1
 1  sin  1 ; x  (5,5)  0 Why ?
x
( Multiplying by x 2 ...Note that x 2 is non  negative)
1
  x 2  x 2 sin  x 2 ; x  (5,5)  0
x
We have'
lim ( x
2
)  0 
x 0
lim ( x
x 0
 By the squeez theorem,
1
2
(
x
sin
)0
lim
x
x 0
2
)
Example (4)
Find
2
4
(
x
cos
)
lim
x
x 0
Soluion :
:
We have :
2
 1  cos  1 ; x  (5,5)  0 Why ?
x
( Multiplying by x 4 ...Note that x 4 is non  negative)
2
  x 4  x 4 cos  x 4 ; x  (5,5)  0
x
We have'
lim ( x
4
)  0 
x 0
lim ( x
4
x 0
 By the squeeze theorem,
2
4
(
x
cos
)0
lim
x
x 0
)
Example (5)
Find


lim  x  7  cos
3
x  7
 

x  7 
Soluion :
:
We have :
 1  cos

x7
 1 ; x  (, )  {7}
and so :
 1  cos 3

x7
 1 ; x  (, )  {7}
( Multiplying by x  7 , which is positive on (, )  {7}
  x  7  x  7 cos 3

x7
 x7
We have'
lim  x  7 
0 
x  7
lim x  7
x  7
 By the squeeze theorem,


lim  x  7  cos
x  7
3
 
 0
x  7 
; x  (, )  {7}