Transcript Limits and Continuity •Definition •Evaluation of Limits •Continuity •Limits Involving Infinity Limit We say that the limit of f ( x) as x approaches a.

Limits and Continuity
•Definition
•Evaluation of Limits
•Continuity
•Limits Involving Infinity
Limit
We say that the limit of f ( x) as x approaches a is L and write
lim f ( x)  L
xa
if the values of f ( x) approach L as x approaches a.
y  f ( x)
L
a
Limits, Graphs, and Calculators
 x 1 
1. a) Use table of values to guess the value of lim  2 
x 1 x  1


x 1
b) Use your calculator to draw the graph f ( x)  2
x 1
and confirm your guess in a)
Graph 1
2. Find the following limits
 sin x 
a) lim 
by considering the values

x 0
 x 
x  1, 0.5, 0.1, 0.05, 0.001. Thus the limit is 1.
sin x
Confirm this by ploting the graph of f ( x) 
x
Graph 2
 x  by considering the values
b) limsin 
x 0
(i) x  1,  1 ,  1
, 1
10
100
1000
(ii) x  1,  2 ,  2
, 2
3
103
1003
This shows the limit does not exist.
 x
Confrim this by ploting the graph of f ( x)  sin 
Graph 3
c) Find
3x if x  2
lim f ( x) where f ( x)  
x2
1 if x  2
lim f ( x) = lim  3 x
x 2
6
 3 lim x
x 2
 3(2)  6
Note: f (-2) = 1
is not involved
x 2
-2
3)
Use your calculator to evaluate the limits
a.
 4( x 2  4) 
lim 

x2
x

2


b.
1, if x  0 Answer : no limit
lim g ( x), where g ( x)  
x 0
1, if x  0
c.
1
lim f ( x), where f ( x)  2
x 0
x
d.
 1 x 1 
lim 

x 0
x


Answer : 16
Answer : no limit
Answer : 1/2
The  - Definition of Limit
We say lim f ( x)  L if and only if
x a
given a positive number  , there exists a positive  such that
if 0 | x  a |  , then | f ( x)  L |  .
L
L 
y  f ( x)
L
a
a 
a 
This means that if we are given a
small interval ( L   , L   ) centered at L,
then we can find a (small) interval (a   , a   )
such that for all x  a in (a   , a   ),
f ( x) is in ( L   , L   ).
Examples
1.
Show that lim(3 x  4)  10.
x2
Let   0 be given. We need to find a   0 such that
if | x - 2 |  , then | (3x  4)  10 |  .
But | (3x  4)  10 || 3x  6 | 3| x  2 | 
if | x  2 |

3
So we choose  

3
.
1
2. Show that lim  1.
x 1 x
Let   0 be given. We need to find a   0 such that
if | x  1|  , then | 1  1|  .
x
x 1 1
1
But |
 1||
| | x  1| .
x
x
x
What do we do with the
x?
1
If we decide | x  1| , then 1  x  3 .
2
2
2
1
And so <2.
x
1
1
Thus |
 1| | x  1| 2 | x  1| .
x
x
 1 
Now we choose   min  ,  .
3 2
1/2 1 3/2
One-Sided Limit
One-Sided Limits
The right-hand limit of f (x), as x approaches a,
equals L
written: lim f ( x)  L
x a
if we can make the value f (x) arbitrarily close
to L by taking x to be sufficiently close to the
right of a.
y  f ( x)
L
a
The left-hand limit of f (x), as x approaches a,
equals M
written:
lim f ( x)  M
x a
if we can make the value f (x) arbitrarily close
to L by taking x to be sufficiently close to the
left of a.
y  f ( x)
M
a
Examples
Examples of One-Sided Limit
1. Given
 x 2 if x  3
f ( x)  
 2x if x  3
Find lim f ( x)
x 3
lim f ( x)  lim 2 x  6
x 3
x 3
Find lim f ( x)
x 3
lim f ( x)  lim x 2  9
x3
x3
More Examples
 x  1, if x  0
2. Let f ( x)  
 x  1, if x  0.
a)
b)
Find the limits:
lim f ( x)  lim ( x  1)  0  1  1
x 0
x 0
lim f ( x)  lim ( x  1)  0  1  1
x 0
x 0
x

1)
c) lim f ( x)  lim(
 11  2

x 1
d) lim f ( x)
x 1
x 1
 lim(
x

1)

x 1
 11  2
A Theorem
lim f ( x)  L if and only if lim f ( x)  L and lim f ( x)  L.
xa
x a
x a
This theorem is used to show a limit does not
exist.
For the function
 x  1, if x  0
f ( x)  
 x  1, if x  0.
lim f ( x) does not exist because lim f ( x)  1 and lim f ( x)  1.
x 0
x 0
x 0
But
lim f ( x)  2 because lim f ( x)  2 and lim f ( x)  2.
x 1
x 1
x 1
Limit Theorems
If c is any number, lim f ( x)  L and lim g ( x)  M , then
x a
a)
c)
e)
g)
i)
x a
b) lim  f ( x)  g ( x)   L  M
lim  f ( x)  g ( x)   L  M
x a
x a
lim  f ( x)  g ( x)   L  M
xa
x a
lim  c  f ( x)   c  L
x a
lim c  c
xa
lim x n  a n
x a

lim f ( x)
d)
f)
h)
j)
 L , ( M  0)

g ( x)
M
lim  f ( x)   Ln
n
x a
lim x  a
xa
lim
x a
f ( x)  L , ( L  0)
Examples Using Limit Rule
Ex. lim  x  1  lim x  lim1
2
2
x3
x3
x3
   lim1
 lim x
2
x 3
x 3
 32  1  10
lim  2 x  1
2 lim x  lim1
2x 1
x 1
x 1
x 1
Ex. lim


x 1 3 x  5
lim  3 x  5 
3lim x  lim 5
x 1
x 1
x 1
2 1 1


35 8
More Examples
1. Suppose lim f ( x)  4 and lim g ( x)  2. Find
x 3
x 3
f ( x)  lim g ( x)
a) lim  f ( x)  g ( x)   lim
x 3
x 3
x 3
 4  (2)  2
b) lim  f ( x)  g ( x)   lim f ( x)  lim g ( x)
x 3
x 3
 4  (2)  6
x 3
2 f ( x)  lim g ( x)
 2 f ( x)  g ( x)  lim
x 3
x 3
c) lim 


x 3
lim f ( x)  lim g ( x)
 f ( x) g ( x) 
x 3
x 3

2  4  (2) 5

4  (2)
4
Indeterminate Forms
Indeterminate forms occur when substitution in the limit
results in 0/0. In such cases either factor or rationalize the
expressions.
Ex.
x5
Notice
x 5 x 2  25
x5
 lim
x5  x  5  x  5 
lim
1
1
 lim

x5  x  5 
10
0
form
0
Factor and cancel
common factors
More Examples
a)
 x 3
lim 
 =
x 9
 x 9 
 ( x  3)( x  3) 
lim 

x 9
 ( x  9)( x  3) 



x9

lim
 lim 

x 9 
x 9 ( x  9)( x  3)



b)
 4  x2 
lim  2 3  = lim  (2  x)(2  x) 
x 2 2 x  x

 x2  x 2 (2  x) 
 2 x 
= lim 

2
x 2
 x 
2  (2) 4

 1
2
(2)
4
1  1
6
x 3
The Squeezing Theorem
If f ( x)  g ( x)  h( x) when x is near a, and if
g ( x)  L
lim f ( x)  lim h( x)  L, then lim
xa
xa
x a
 
Example: Show that lim x2 sin   0.
x
x 0
Note that we cannot use product rule because lim sin 
x 0
 
 
But  1  sin 
 x  DNE!
 1 and so  x 2  x 2 sin   x 2 .
x
x
Since lim x 2  lim(  x 2 )  0, we use the Squeezing Theorem to conclude
x 0
x 0
 x   0.
lim x2 sin 
x 0
See Graph
Continuity
A function f is continuous at the point x = a if
the following are true:
i) f (a) is defined
ii ) lim f ( x) exists
x a
f(a)
a
A function f is continuous at the point x = a if
the following are true:
i) f (a) is defined
ii ) lim f ( x) exists
x a
iii ) lim f ( x)  f (a )
xa
f(a)
a
Examples
At which value(s) of x is the given function
discontinuous?
2
x 9
1. f ( x)  x  2
2.
g ( x) 
x3
Continuous everywhere
Continuous everywhere
except at x  3
lim( x  2)  a  2
x a
and so lim f ( x)  f (a)
x a
g (3) is undefined
4
6
2
4
-6
-4
-2
2
-2
2
-4
-4
-2
2
-2
4
-6
-8
-10
4
1, if x  0
4. F ( x)  
1, if x  0
 x  2, if x  1
3. h( x)  
1, if x  1
lim h( x)  1 and lim h( x)  3
x 1
x 1
Thus h is not cont. at x=1.
h is continuous everywhere else
lim F ( x)  1 and lim F ( x)  1
x 0
x 0
Thus F is not cont. at x  0.
F is continuous everywhere else
5
3
4
2
3
2
1
1
-10
-2
2
4
-5
5
-1
-1
-2
-3
-2
-3
10
Continuous Functions
If f and g are continuous at x = a, then
f  g , fg , and f
g
 g (a)  0  are continuous
at x  a
A polynomial function y = P(x) is continuous at
every point x.
A rational function R( x)  p( x) q( x) is continuous
at every point x in its domain.
Intermediate Value Theorem
If f is a continuous function on a closed interval [a, b]
and L is any number between f (a) and f (b), then there
is at least one number c in [a, b] such that f(c) = L.
y  f ( x)
f (b)
f (c) = L
f (a)
a c
b
Example
Given f ( x)  3x  2 x  5,
2
Show that f ( x)  0 has a solution on 1, 2.
f (1)  4  0
f (2)  3  0
f (x) is continuous (polynomial) and since f (1) < 0
and f (2) > 0, by the Intermediate Value Theorem
there exists a c on [1, 2] such that f (c) = 0.
Limits at Infinity
1
1
For all n > 0, lim n  lim n  0
x  x
x  x
1
provided that n is defined.
x
5  1
3

2
3x  5 x  1
x
x
 lim
Ex. xlim
2
2 4
x 

2  4x
x2
2

 x   lim  1 x   3  0  0   3
lim 3  lim 5
x 
x 
 x
lim 2
x 
Divide
2
by x
2
x 
 lim 4
x 
2
04
4
More Examples
 2 x3  3x 2  2 
1. lim  3

x  x  x 2  100 x  1


 2 x3 3x 2 2
 3  3

3
x
 lim  3 x 2 x
x  x
  x  100 x  1
 3
3
3
3
x
x
x
x
3 2


 2  x  x3

 lim 
x 
1 100 1 
 1  2  3 
x x
x 

2
 2
1






 4 x  5 x  21 
lim  3

x  7 x  5 x 2  10 x  1


2
2.


4 x 2 5 x 21
 3 3


3
x
x
x
 lim  3

2
x  7 x
5 x 10 x 1 

 3  3  3  3
x
x
x 
 x
 4 5 21
 x  x 2  x3
 lim 
x 
 7  5  102  13
x x
x

0

7
0





3.
 x2  2 x  4 
lim 

x 
 12 x  31 
 x2 2x 4 
 
 
x x
 lim  x
x 
 12 x  31 
 x

x


4

 x2 x 
 lim 
x 
31 
 12 

x 

2

12

4.

 lim 
x 


lim
x 


x2  1  x
x2  1  x
1



x 1  x 

x2  1  x 

2
 x2  1  x2 
 lim 

2
x 
 x 1  x 


1
 lim 

2
x 
 x 1  x 
1
1

 0
 
Infinite Limits
20
For all n > 0,
15
10
lim
xa
1
 x  a
n
5

-8
-6
-4
-2
2
-5
-10
-15
-20
40
lim
x a
30
1
 x  a
n
  if n is even
20
10
-2
2
4
6
-10
-20
lim
x a
20
1
 x  a
n
  if n is odd
15
10
5
-8
-6
-4
-2
2
-5
More Graphs
-10
-15
Examples
Find the limits
1.
2.
 3x 2  2 x  1 
lim 

2
x 0
2x


 3 2  1 2 
x
x 
= lim 

x 0 
2



 2x 1 
 2x  1 
lim 
 = lim
x 3  2 x  6  x3  2( x  3) 


40
20
-8
-6
-4
-2
2
-20
3   


2
 
Limit and Trig Functions
From the graph of trigs functions
f ( x)  sin x and g ( x)  cos x
1
1
0.5
0.5
-10
-5
5
10
-10
-0.5
-5
5
10
-0.5
-1
-1
we conclude that they are continuous everywhere
lim sin x  sin c and lim cos x  cos c
x c
x c
Tangent and Secant
Tangent and secant are continuous everywhere in their
domain, which is the set of all real numbers
x  
2
,  3
2
,  5
2
,  7
2
y  sec x
,
15
30
y  tan x
10
20
5
10
-6
-6
-4
-2
2
4
6
-4
-2
2
-5
-10
-10
-20
-15
-30
4
6
Examples
a) lim  sec x 
 2
x 
c)

lim  tan x  

x  3
2

e) lim cot x  
x 
b)
d)
lim  sec x
 2
x 

lim  tan x 

x  3
2

f) lim tan x  1
x 
4
cos x 0
 0
g) lim cot x  xlim
3  sin x
1

x  3 
2
2

Limit and Exponential Functions
y  ax , a  1
10
10
y  ax , 0  a  1
8
8
6
6
4
4
2
2
-6
-4
-2
2
4
-6
6
-4
-2
2
4
-2
-2
The above graph confirm that exponential
functions are continuous everywhere.
lim a  a
x
x c
c
6
Asymptotes
The line y  L is called a horizontal asymptote
of the curve y  f ( x) if eihter
lim f ( x)  L or lim f ( x)  L.
x 
x 
The line x  c is called a vertical asymptote
of the curve y  f ( x) if eihter
lim f ( x)   or lim f ( x)  .
x c
x c
Examples
Find the asymptotes of the graphs of the functions
x 1
1. f ( x)  2
x 1
(i) lim f ( x)  
2
x 1
Therefore the line x  1
is a vertical asymptote.
(ii) lim f ( x)  .
x 1
Therefore the line x  1
is a vertical asymptote.
(iii) lim f ( x)  1.
x 
Therefore the line y  1
is a horizonatl asymptote.
10
7.5
5
2.5
-4
-2
2
-2.5
-5
-7.5
-10
4
2.
x 1
f ( x)  2
x 1
 x 1 
(i) lim f ( x)  lim  2 
x 1
x 1
 x 1 
(iii) lim f ( x)  0.
x 
Therefore the line y  0


x 1
 1  1
= lim 
 lim 
 . is a horizonatl asymptote.


x 1
 ( x  1)( x  1)  x1  x  1  2
Therefore the line x  1
is NOT a vertical asymptote.
(ii) lim f ( x)  .
x 1
Therefore the line x  1
is a vertical asymptote.
10
7.5
5
2.5
-4
-2
2
-2.5
-5
-7.5
-10
4