SECTION 1.4:
CONTINUITY AND ONE-SIDED LIMITS
Goals: The Student Will Be Able To:

Define and apply the concepts of Continuity at a
Point, Continuity on an Open Interval and
Continuity on a Closed Interval

Recognize, understand and distinguish between
Nonremovable and Removable discontinuities

Discuss the continuity of various functions, and
corresponding graphical implications

Differentiate between infinite and jump
Nonremovable discontinuities

Evaluate limits from the left and right (one-sided
limits)

Understand, explain and apply the Intermediate
Value Theorem
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DEFINITION OF CONTINUITY
Continuity at a Point:
A function is continuous at “c” if each of t he these
three (3) conditions are met:
1. f (c) is defined
lim f ( x)
2. xc
exists
lim f ( x)  f (c)
3. xc
Continuity on an Open Interval:
A function is continuous on an open interval (a, b), if it
is continuous at every point within the interval.
Continuity Everywhere:
A function is continuous everywhere, if (a, b) is (, ) .
2
EXAMPLES OF DISCONTINUITY
Rule 2:
f(c) is defined, but
left and right limits
do not match …so
limit does not exist
Rule 1:
f(c) is not defined
(
a
c
)
b
(
a
c
)
b
Rule 3:
left and right limits
do match …f(c) is
defined, but f(c) 
limit
(
a
c
)
b
3
2 TYPES OF DISCONTINUITYREMOVABLE AND NONREMOVABLE
REMOVABLE: If the discontinuity can be taken away
by defining or redefining a value for the function. “Fill
in the hole, or erase the point and move it to the hole.”
(
a
c
)
b
(
a
c
)
b
NONREMOVABLE: If the discontinuity cannot be
taken away. Defining or redifinig a point will not
reconnect the function. “JUMP” and “INFINITE”
discontinuities are two basic types.
INFINITE
JUMP
(
a
c
)
b
c
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EXAMPLE 1:
Discuss the continuity of each function:
1
f ( x) 
a.
x
Since at x = 0, the function is not defined, there is a
discontinuity at x = 0. Therefore f is continuous for
(,0)  (0, ) . Considering the familiar graph we can
see it is a NONREMOVABLE – INFINTE type.
b.
x2 1
f ( x) 
x 1
Again, we can see that x =1 will produce a 0 in the
denominator. All other x values are in the domain, so f
is continuous for (,1)  (1, ) . However, due to
analytic cancellation, g ( x)  x  1 is a function identical
to f except for the “hole” at (1, 2). Since the
discontinuity is just a “hole”, this is the REMOVABLE
type.
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 x 1 x  0
f ( x)   2
c.
x  1 x  0
This is a special type of function which joins more than
1 equation. 2 separate pieces of graphs are used. f is
called a “PECEWISE” function. Since thepieces join,
and do not “JUMP”, the function is CONTINUOUS for
all x … (, ) .
d. f ( x)  sin x
The domain for this trigonometric function is all real
numbers. There are no discontinuities at all. So the
interval of continuity is (, ) .
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ONE-SIDED LIMITS
We have already suggested that it is necessary to
examine a neighborhood both to the left and to the right
of “c”, when evaluating a limit. Now we will give it a
formal name and formal notation.
lim f ( x)
“Limit from the right”  xc
Examine the function for values greater than c … or
the “positive side of” c.
lim f ( x)
“Limit from the left”  xc
Examine the function for values less than c … or the
“negative side of” c.
EXAMPLE 2:
2
f
(
x
)

4

x
Find the limit of
as x approaches –2 from
the right.
With either numerical analysis, or
simple visual inspection we can see
the function is approaching 0, as x
approaches –2 from the right.
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.EXAMPLE 3:
Find the limit of the “Greatest Integer Function”
f ( x)  [[ x]] as x approaches 0 from the left and from the
right.
Since [[0.1]]  0 , or no matter how close x gets to 0 from
lim [[x]]  0
the right, f is still 0 … x0
Yet as x gets close to 0 from the left,
The value of the function remains
a steady –1. Ex: [[0.1]]  1
lim [[x]]  1
So then x0
We can conclude that the “Greatest Integer Function” is
not continuous at ANY integer value. These are all
JUMP discontinuities
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THE EXISTENCE OF A LIMIT
Let f be a function and let c and L be real numbers. The
limit as x approaches c exists if and only if:
lim f ( x)  L
xc 
and
lim f ( x)  L
xc 
CONTINUITY ON A CLOSED INTERVAL
A function f is continuous on an closed interval [a, b],
if it is continuous on the open interval (a, b) and:
lim f ( x)  f (a)
xa 
and
lim f ( x)  f (b)
xb
Basically, the endpoints need to fill filled-in pieces of
the graph.
9
2
f
(
x
)

1

x
EXAMPLE 4: Discuss the continuity of
For this radical function, it is clearly
continuous on the open interval (-1, 1).
And since
lim f ( x)  f (1)  0
x1
lim f ( x)  f (1)  0
x1
… Continuous from the right at –1
… Continuous from the left at 1
We can conclude that f is continuous on the closed interval
[-1, 1], as this can be visually verified on the graph.
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PROPERTIES OF CONTINUITY
If b is a real number and f and g are continuous at x = c,
then the following functions are also continuous at x = c.
1. Scalar Multiple: bf
2. Sum and Difference: f  g
3. Product: f  g
4. Product: f / g, g  0
Two continuous functions can be multiplies by a constant,
added, subtracted, multiplied or divided by each other, and
still be continuous!
EXAMPLE 7: Describe the intervals on which each
function is continuous.

a. f ( x)  tan x … The tangent function is undefined at 2 ,
and at repeated intervals n where n is an integer. It is
continuous within these discontinuous values. So it is
continuous on open intervals:
 3         3 
... 
,     ,    , ...
2
2  2 2 2 2 

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THE INTERMEDIATE VALUE THEOREM
If f is continuous on the closed interval [a, b], and k is
any number between f(a) and f(b), then: the exists () at
least one number c in [a, b] such that f(c) = k.
f(a)
f(a)
k
k
f(b)
f(b)
[ c1
a
c2
c3 ]
b
2
Theorem Applies
[
a
]
b
Theorem Doesn’t Apply
For any theorem to apply, the hypothesis must be true. So to
apply the IVT, we must verify/state the truth of the 2 parts of
the hypothesis.
1. The function must be continuous on [a, b]
2. k is selected between the two y values … f(a) and f(b)
 3. Conclusion: There is at least one c value within (a,
b) such that f(c) = k. We show 3 c’s in our example above.
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EXAMPLE 8:
Use the IVT to show that the function
f ( x)  x 3  2x  1 has a zero in the interval [0, 1].
Solution: To apply the IVT, recall, we must establish
the 2-part hypothesis is valid.
1. Since f is a polynomial, it is permissible to
assume that it is continuous for all x, and therefore on
our interval [0, 1]
2. f(0) = -1 and f(1) = 2, an the k value suggested in
the problem is Zero. It is true that 0 is between the two y
values –1 and 2.
So, the IVT will apply, and there must be at least 1 c
value between 0 & 1 which will yield f(c) = 0
A graphing calculator can be used
to find that “c” … the IVT just
guarantees its existence. Also, the
graph to the right illustrates the
value.
k
c
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