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Continuity and One-Sided Limits
Lesson 1.4
Calculus
• 1.4 Continuity and 1-sided limits
• Student objectives:
– Understand & describe & find continuity at a
point vs. continuity on an open interval
– Find 1-sided limits
– Use properties of continuity
– Understand & use the Intermediate Value
Theorem
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Question??
• How do I get from point A to point B?
School
Home
3
Intuitively, a function is continuous at x = c if you
can draw it without lifting your pen from the paper.
In the diagram below, the function on the left is
continuous throughout, but the function on the
right is not. It is "discontinuous" at x = c.
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Formally, a function is continuous at x = c on
the open interval (a, b) if
the following three conditions are met:
f(c) is defined
lim f ( x ) exists
xc
lim f ( x)  f (c)
x c
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1.4: Examples of Discontinuity
• F is continuous at x=c if there are no holes,
jumps or gaps at c
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Graphically:
f (x) is continuous at c.
lim f ( x)  f (c)
x c
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1.4: Removable and
Nonremovable discontinuity
Removable
discontinuity
Nonremovable
discontinuity
Removable
discontinuity
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Discuss the continuity of the following graph.
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Continuity at a Point
• A function can be discontinuous at a point
– The function jumps to a different value at a
point
– The function goes to infinity at one or both
sides of the point
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Discuss the continuity of each of
the following functions:
1
f ( x) 
x
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The domain of f is all nonzero real numbers.
You can conclude that f is continuous at every
x-value in its domain. In other words, there
is no way to define f(0) so as to make the
function continuous at x = 0.
1
f ( x) 
x
Nonremovable Discontinuity
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Discuss the continuity of each of
the following functions:
x2  1
f ( x) 
x 1
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The domain of f is all real numbers except x = 1.
Therefore, you can conclude that f is
continuous at every x-value in its domain.
At x = 1, the function has a removable discontinuity.
If f(1) is defined as 2, the “newly defined” function
is continuous for all real numbers.
x 1
f ( x) 
x 1
2
Removable Discontinuity
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Discuss the continuity of each of
the following functions:
f ( x)  sin x
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The domain of f(x) is all real numbers.
Therefore, you can conclude that the function is
continuous on its entire domain.
When a function is continuous from (-∞,∞),
we say that the function is everywhere continuous.
f ( x)  sin x
Everywhere Continuous
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Which of These is Dis/Continuous?
• When x = 1 … why or not
x2  2 x  3
h( x ) 
x 1
if x ≠ 1 and
h(x) = 4 if x = 1
x2  2x  3
f ( x) 
x 1
x2  2 x  3
g ( x) 
x 1
x3
F ( x) 
x 1
if x ≠ 1 and
g(x) = 6 if x = 1
if x ≠ 1 and
F(x) = 4 if x = 1
Are any removable?
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Continuity Theorem
• A function will be continuous at any
number
x = c for which f(c) is defined, when …





f(x) is a polynomial
f(x) is a power function
f(x) is a rational function
f(x) is a trigonometric function
f(x) is an inverse trigonometric function
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Theorem 1.11 Properties of
Continuity
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Theorem 1.12 Continuity of a
Composite Function
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Removing a discontinuity:
x3  1
f  x  2
x 1
has a discontinuity at x  1 .
Write an extended function that is continuous
at x  1 .
 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
 x 2  1 , x  1
f  x  
 3 , x 1
 2

3
2
Note: There is another
discontinuity at x  1 that can
not be removed.
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Make sure you have all this
down
lim f ( x )
x b
lim f ( x )
x b
A plus sign means from the right and a
minus sign means from the left
If the limit exists then the limit is the same
from the left and from the right.
A table can be helpful when you can’t find
a general limit.
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Theorem 1.10 The Existence of
a Limit
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Left & Right Hand Limits
lim f ( x)
lim f ( x)
x a
x a
2
-5
lim x 3  0
x 0
5
-2
lim x 3  0
x 0
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Theorem
10
50
40
5
30
-10
-5
0
-5
-10
5
x
The limit does
not exist for
this function.
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The limit exists
for this function.
20
10
0
5
10
x
15
20
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lim f ( x)  2
x 1
The limit is
a number
lim f ( x)  2
x 1
lim f ( x)  2
x 1
The limit can exist even
when the function is not
defined at a point or has a
value different from the limit.
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Section 1.4: Continuity
Rationals
• Continuity of a closed
interval:
lim 1  x 2  0  f (1)
• Right Cont. x1
lim 1  x 2  0  f (1)
• Left Cont. x1
• Conclude that f is
continuous on the
closed interval
1,1

1
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Greatest Integer Function:
y  greatest integer that is  x
x
y
0
0.5
0.75
1
1.5
2
0
0
0
1
1
2
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Greatest Integer Function:
y  greatest integer that is  x
The greatest integer function is
also called the floor function.
The notation for the floor function
is:
y x
Find its limit as x approaches 0
from the right and left!!!
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
lim
x 3
x 3
0
D.S. 
0
x 3
Use piecewise definition for absolute value functions
 x x0
x 
 x x  0
 x 3 x 3 0  x  3
 x 3  
 ( x  3) x  3  0  x  3
x 3
x 3
lim

1
x 3 x  3
x 3
x 3
 ( x  3)
lim

 1
x 3 x  3
x 3
Limit DNE!
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 x2 1 x  2
Determine lim f ( x) if f ( x)   2
x 2
 x 5 x  2
lim x  1  2  1  3
2
2
x 2
lim x  5  2  5  9  3
2
2
x 2
 lim f ( x)  3
x 2
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Properties of Continuous Functions
• If f and g are functions, continuous at x = c
Then …
 s  f ( x ) is continuous (where s is a constant)


f(x) + g(x) is continuous
f ( x)  g ( x) is continuous

f ( x)
is continuous
g ( x)

f(g(x)) is continuous
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One Sided Continuity
• A function is continuous from the right at a
point x = a if and only if lim f ( x)  f (a)
x a
• A function is continuous from the left at a
f ( x)  f (b)
point x = b if and only if xlim
b

a
b
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Continuity on an Interval
• The function f is said to be continuous on
an open interval (a, b) if
– It is continuous at each number/point of the
interval
• It is said to be continuous on a closed
interval [a, b] if
– It is continuous at each number/point of the
interval and
– It is continuous from the right at a and
continuous from the left at b
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Continuity on an Interval
• On what intervals are the following
functions continuous?
g ( x)  sin x
x2  1
f ( x)  2
x 4
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Intermediate Value Theorem
If a function is continuous between a and b, then it takes
on every value between f  a  and f  b  .
f b
Because the function is
continuous, it must take on
every y value between f  a 
and f  b  .
f a
a
b
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
Locating Roots with Intermediate
Value Theorem
• Given f (a) and f (b) have opposite sign
– One negative, the other positive
• Then there must be a root between a and b
a
b
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Intermediate value theorem, bounds.
Intermediate value theorem: Given a continuous function
in the interval [a,b], if f(a) and f(b) are of different signs,
then there is at least one zero between a and b.
Given f(x) is continuous , in what interval
does a zero exist? Explain.
f ( x)  x 2  5 x  3
a ) [3,4]
b) [4,5]
c) [5,6]
f(3) = -9
f(4) = -7
f(5) = -3
There is a zero in the interval [5,6] because
there is a sign change, and by intermediate
value theorem, a zero must exist in that interval.
f(6) = 3
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Discuss the continuity of each.
3 x 2 for x  2
1. f ( x )  
ax for x  2
Find a so that f ( x ) is a continuous function.
Discuss the continuity of each.
3 x 2 for x  2
1. f ( x )  
ax for x  2
Find a so that f ( x ) is a continuous function.
If x < 2, the function is a parabola. (continuous)
If x > 2, the function is a line. (continuous)
To be continuous, the two sides must also meet when x = 2.
lim  f ( x )  lim  3 x 2   12
x 2
lim  f ( x ) 
x 2
x 2
lim  ax   2a
x 2
2a  12
a6
Intermediate Value Theorem
If f is continuous on [a,b] and k is any number between f(a) and
f(b), then there exists a number c in [a,b] such that f(c) = k.
4
2
-5
5
-2
-4
Intermediate Value Theorem
If f is continuous on [a,b] and k is any number between f(a) and f(b),
then there exists a number c in [a,b] such that f(c) = k.
4
2
a
-5
b
c
c
-2
k
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The red graph has 1 c-value.
Orange has 1 c-value.
-4
Blue has 5 c-values.
Translation:
If you connect two dots with a continuous function, you must hit every y-value
between them at least once.
Removable Discontinuities:
(You can fill the hole.)
Essential Discontinuities:
jump
infinite
oscillating
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