Continuity
Section 2.3
Calculus
Warm-Up: Find each limit
a.
 4( x 2  4) 
lim 

x 2
x

2


Answer : 16
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 : no limit
Answer : 1/2
More Examples
 x  1, if x  0
2. Let f ( x)  
 x  1, if x  0.
a)
b)
lim f ( x)  lim ( x  1)
x 0
x 0
 0 1  1
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
Examples Using Limit Rule
 
x  lim1  lim x  lim1
Ex. lim  x  1  lim
x 3
x 3
2
x3
2
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
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)
1  1
 6
x 3
 4  x2 
 (2  x)(2  x) 
lim  2
 = lim 
x 2 2 x  x3

 x2  x 2 (2  x) 
 2 x
= lim 

2
x 2
 x

2  (2) 4

 1
2
(2)
4
Examples
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
 
2.3 Continuity
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
Most of the techniques of calculus require that functions be continuous. A
function is continuous if you can draw it in one motion without picking up your
pencil.
A function is continuous at a point if the limit is the same as the value of the
function.
This function has discontinuities at x=1 and
x=2.
2
1
1
2
3
4
It is continuous at x=0 and x=4, because
the one-sided limits match the value of the
function

Removable Discontinuities:
(You can fill the hole.)
Essential Discontinuities:
jump
infinite
oscillating

Removing a discontinuity:
 x  1  x 2  x  1 1  1  1
x3  1
 lim

lim 2
x 1
x 1 x  1
 x  1 x  1
2
 x3  1
 2 , x  1
f  x    x 1
 3 , x 1
 2
3

2
Removing a discontinuity:
5
4
3
2
1
-5 -4 -3 -2 -1 0
-1
1
2
3
4
5
-2
-3
-4
-5
 x3  1
 2 , x  1
f  x    x 1
 3 , x 1
 2
Note: There is another discontinuity at
that can not be removed.
x  1

Use the function below to answer the
questions below:
1.) At which points is
f(x) discontinuous?
2.) What type of
discontinuity at each
point?
Use the function below to answer the
questions below:
1.) At which points is
f(x) discontinuous?
2.) What type of
discontinuity at each
point?
Intermediate Value Theorem
If a function is continuous between
between
f a
and
f b .
a and b, then it takes on every value
f b
Because the function is continuous, it
must take on every y value between
.
f b and
 
f a
a
f a
b

Example 5:
Is any real number exactly one less than its cube?
(Note that this doesn’t ask what the number is, only if it exists.)
f 1  1
x  x3  1
0  x3  x  1
f  x   x3  x  1
f  2  5
Since f is a continuous function, by the intermediate
value theorem it must take on every value between 1 and 5.
Therefore there must be at least one solution
between 1 and 2.
Use your calculator to find an approximate solution.
solve  x  x3  1, x 
F2
1: solve
1.32472

Find the constant “a” and “b” such that the
function is continuous on the entire real
line.
ì2
x < -1
ï
f (x) = í ax + b
-1 £ x £ 3
ï-2
x>3
î
(-1, 2)
Find the slope between
those two points
m=
2 - (-2)
= -1
-1- 3
We need to connect the
dots with a line, including
the dots, so now we know
the slope, just figure out
what the y-intercept is.
(0, 1)
a = -1
b=1
(3, -2)
Find the constant “a” such that the
function is continuous on the entire real
line.
ìï x3
f (x) = í 2
ïî ax
Find the limits from the
left and right
lim- x3 = 8
x®2
lim+ ax2 = 4a
x®2
8 = 4a, therefore a = 2
x£2
x>2
Determine if the discontinuity is removable or non-removable.
c) Graph first!!
ì-3x if x ¹ -2
f (x) = í
î1 if x = -2
6
f(x) has
removable
discontinuity
at x = -2
-2
Example 1 INVESTIGATING CONTINUIT Y
  x  1,
 1,

f ( x )   2,
 x  1,
 x  5,

3
0  x 1
1 x  2
x 2
2x 3
3x 4
Solution
This function f is continuous
2
at every point in its domain
[0, 4], except at x  1 and
1
x  2.
0
0
1
2
3
4
5
Example 1 INVESTIGATING CONTINUITY
Solution
point at which f is continuous
x  0,
lim f (x )  f (0)
x 0
x  4,
lim f (x )  f (4)
x 4
x  c , c  1, 2, lim f (x )  f (c )
3
point at which f is discontinuous
2
x c
x  1,
x  2,
lim f ( x )  does not exist
x 1
lim f ( x )  1, but 1  f (2)
1
x 2
0
0
1
2
3
4
5
1
y = f(x)
y = f(x)
1
continuous at x=0
1
2
y = f(x)
continuous at x=0
If it had f(0)=1
continuous at x=0
If it had f(0)=1
continuity at x = 0 are
removable
lim f ( x ) does not exist, no way to
x 0
1
improve it. It is a trype of a jump
y = f(x)
discontinuity.
6
lim1/ x 2  ,
4
x 0
an infinite discontinuity.
2
0
-4
-2
0
2
4
Exploration 1 Removing a Discontinuity
Consider the function
x 3  7x  6
f (x ) 
x2 9
1. Factor the denominator. What is the domain of f (x )
2. Investigate the graph of f around x  3 to see that f
has a removable discontinuity at x  3
40
0
-4
-2
0
-40
2
4
Example 3 IDENTIFYING CONTINUOUS FUNCTIONS
Consider the function
f ( x )  1/ x
Solution
It is a continuous function because it is continuous at every
point of its domain. However, it has a point of discontinuity
at x  0 because it is not defined there.
4
2
0
-4
-2
-2
-4
0
2
4
Examples
At which value(s) of x is the given
function discontinuous?
2
x 9
f ( x)  x  2
2.
g ( x) 
x3
1.
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
and
lim F ( x)  1
lim h( x)  1 and lim h( x)  3
x 1
x 1
x 0

Thus F is not cont. at
Thus h is not cont. at
x=1.
lim F ( x)  1
x  0
x  0.
F is continuous everywhere else
h 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