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2.3 Continuity
Grand Canyon, Arizona
Photo by Vickie Kelly, 2002
Greg Kelly, Hanford High School, Richland, Washington
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

Where have we seen those types of
discontinuities before?
A. They were in a movie we watched in class.
B. They were topics of McKenzie’s bad jokes.
C. They were the reasons for limits not existing.
D. They were the names of three of the seven
dwarfs.
We said this function has a ‘removable discontinuity’,
how can we also describe it with ‘Calculus terms”?
A. It has a derivative at a.
B. It has a limit at a.
C. The integral exists at a.
D. It is continuous at a.
a
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
 2 , x  1
f  x    x 1
 3 , x 1
 2
3

2
Note: There is another
discontinuity at x  1 that can
not be removed.
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 x  1 that can
not be removed.

Continuous functions can be added, subtracted, multiplied,
divided and multiplied by a constant, and the new function
remains continuous.
Also: Composites of continuous functions are continuous.
examples:
y  sin  x 2 
y  cos x

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

If we have a continuous function that has f(3)=-2
and f(4)=4, which of these MUST be true?
A. f(5) must be positive
B. f(2) must be negative
C. f(x)=0 must have a solution between 3 and 4
D. f(x) must be a linear equation with a slope of 6
E. None of the above
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

Graphing calculators can sometimes make noncontinuous functions appear continuous.
Graph:
y  floor  x 
CATALOG
Note resolution.
F
floor(
This example was graphed
on the classic TI-89. You
can not change the
resolution on the Titanium
Edition.
The calculator “connects the dots”
which covers up the discontinuities. 
Graphing calculators can make non-continuous
functions appear continuous.
Graph:
y  floor  x 
CATALOG
F
floor(
If we change the plot style
to “dot” and the resolution
to 1, then we get a graph
that is closer to the
correct floor graph.
The open and closed circles do not
show, but weGRAPH
can see the
discontinuities.

How do we prove that a function is
continuous at a point
Let’s prove that sin(x) is continuous at 𝜋

First things first… what is the first
thing we must know?
A. 𝑓 𝜋 = 0
B. 𝑓 0 = 𝜋
C. lim 𝑓 𝑥 = 0
𝑥→𝜋
D. lim 𝑓 𝑥 = 𝜋
𝑥→0
E. f(x) is a trig funct.

What must we know to prove that
lim 𝑓 𝑥 = 0?
𝑥→𝜋
A. lim− 𝑓(𝑥) = lim
𝑓(𝑥)
+
𝑥→0
𝑥→0
B. lim− 𝑓(𝑥) = lim
𝑓(𝑥)
+
𝑥→𝜋
𝑥→𝜋
C. 𝑓 0 = 𝜋
D. 𝑓 𝜋 = 0

Ok, so if we know that
lim− 𝑓(𝑥) = lim + 𝑓(𝑥)
𝑥→𝜋
𝑥→𝜋
what else do we need to know?
• The limit above tells
us that the function
approaches the same
value from both
sides, what else is
needed?

Looking at the picture, what do we
need to add for f(x) to be continuous?
A. 𝑓 𝜋 = 0
B. 𝑓 0 = 𝜋
C. 𝑓
𝜋
2
=1
D. 𝑓 0 = 0

So to prove continuity at x=c we have to:
1. Prove that the limit exists… so we know that the
function approaches the same value from both
values larger and values lower than the value we
want to prove continuity for.
lim− 𝑓(𝑥) = lim+ 𝑓(𝑥)
𝑥→𝑐
𝑥→𝑐
2. Next, we have to prove that the function has the
same value as the limit. What this means is that we
fill in the hole left by the limit.
lim− 𝑓(𝑥) = lim+ 𝑓 𝑥 = 𝑓(𝑐)
𝑥→𝑐
𝑥→𝑐
p