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More on Continuity
Quick Review
Continuity Checklist. A function f is continuous at a if the following three condi-
tions hold:
1. f ( a ) is defined (i.e., a is in the domain of f ).
2. lim
x! a
f ( x ) exists.
3. lim
x! a
f ( x ) = f ( a ).
Remember:
• A polynomial is continuous everywhere, i.e. for all x.
• A rational functions r ( x ) =
p ( x )
q ( x )
where p and q are polynomials is continuous at
all points in its domain, i.e., where q ( x ) 6 = 0.
a
a
a
Figure 9. 4 : Three of the ways in which
a function f may fail to be continuous
because lim
x! a
f ( x ) 6 = f ( a ).
12. f ( a ) may not be defined;
13. lim
x! a
f ( x ) may not exist;
14. lim
x! a
f ( x ) and f ( a ) may both exist but
not be equal.
EXAMPLE 9. 7. Determine the points at which f ( x ) =
2
x
2
5
x 5
is discontinuous. At
which points does f have VA’s? Removable discontinuities?
SOLUTION. Since f ( x ) is rational it is continuous at all points in its domain. We
can see immediately that f is not defined at x = 5 and x = 0, where there would be
division by 0, so f is not continuous at these two points. Let’s simplify the expression
for f before taking the appropriate limits.
f ( x ) =
2
x
2
5
x 5
=
10 2 x
5 x
x 5
=
10 2 x
5 x ( x 5 )
.
At x = 0:
lim
x! 0
f ( x ) = lim
x! 0
10 2 x
5 x ( x 5 )
= lim
x! 0
2 ( x 5 )
5 x ( x 5 )
= lim
x! 0
2
5 x
|{z}
5 · 0
= 0
= •.
This is enough to conclude that f has a VA at 0.
At x = 5: Having seen the factorization of f , we know that we can calculate a
two-sided limit at 5.
lim
x! 5
f ( x ) = lim
x! 5
2 ( x 5 )
5 x ( x 5 )
= lim
x! 5
2
5 x
Rat’l
=
2
25
.
Since lim
x! 5
f ( x ) exists but f ( 5 ) is not defined, then f has a removable discontinuity at
9
Basic Continuity Properties
One of the most important facts about continuity is that it is preserved under the
standard mathematical operations on functions.
THEOREM 9. 2. (Some Types of Continuous Functions) Assume that f and g are both contin-
uous at x = a and that c is a constant. Then the following new functions are continuous at
x = a.
(a) The sum and differences: f + g and f g
(b) Constant multiples: c f
(c) Products: f g
(d) Quotients:
f
g
, provided g ( a ) 6 = 0
(e) Powers: [ f ( x )]
n
, where n is a positive integer.
Proof. The proofs of all of these follow from the corresponding basic limit prop-
erties. Let’s prove (c). Since f and g are continuous, by Definition 8. 1 this means
lim
x! a
f ( x ) = f ( a ) and lim
x! a
g ( x ) = g ( a ). But then by the product limit property,
lim
x! a
[ f ( x ) g ( x )]
Product
= lim
x! a
f ( x ) · lim
x! a
g ( x )
f , g cont
= f ( a ) g ( a )
so f g is continuous at a. The other parts can be proven in a similar fashion.
YOU TRY IT 9. 5. Try proving part (e). Which limit property must you use?
YOU TRY IT 9. 6 (Think it through). Earlier we the basic limit properties we were able to show
earlier that polynomials were continuous. INstead we could have waited until now to show
polynomials are continuous using the continuity properties in Theorem 9. 2. We know
that f ( x ) = x is a continuous function. Do you see how you prove that x
2
is continuous
everywhere directly from Theorem 9. 2? What parts of Theorem 9. 2 would you need to
use to prove that 2x
2
- 3 x is continuous everywhere? How about a general polynomial
p ( x ) = a n
x
n
1
x + a
0
? Once you know that polynomials are continuous, which prop-
erty in Theorem 9. 2 tells you that rational functions are continuous on their domains?
Composition and Continuity
So far we have not mentioned the operation of composition. This is the most com-
plex of the basic function operations, nonetheless it preserves continuity under the
appropriate circumstances. Functions, like numbers, may be combined using sums,
differences, products, or quotients as we have seen in our limit properties. The
process called composition also produces new functions and this process is unique
to functions and is not possible to carry out on numbers. Composition of func-
tions will play an important role in our development of the calculus. Recall how
composition works.
DEFINITION (Composite Functions). Given two functions f and g, the composite function
f g is defined by ( f g )( x ) = f ( g ( x )). It is evaluated in two steps: y = f ( u ) where
u = g ( x ). The domain of f g consists of all x in the domain of g such that u = g ( x ) is in
the domain of f.
EXAMPLE 9. 8. Let f ( x ) = 3 x
2
x, g ( x ) = x
2
1, h ( x ) =
p
x, and j ( x ) =
1
x
. Simplify
the following expressions. Determine their domains.
(a) ( f g )( x ) (b) ( g f )( x ) (c) ( h f )( x )
(d) ( g g )( x ) (e) ( j j )( x ) (f ) ( f j )( x )
11
Figure 9. 5 : f is left-continuous at x = 1
and right-continuous at x = 9.
Let’s get back to intervals Now we combine this definition with the definition of
continuity at a point (Definition 8. 1 ) to define continuity on an interval.
DEFINITION 9. 4. (Continuity on an Interval) A function f is continuous on an interval I if
f is continuous at all the points of I. If I contains its endpoints, then continuity on I means
left- or right-continuous the right or left endpoints, respectively.
YOU TRY IT 9. 7. Use Figure 9. 5 to answer these questions. Notice that f is continuous on the
interval ( 3, 1 ] since it is continuous from the left at 1 but not from the right at 3. Further
f is not continuous a larger interval containing this one since f is not continuous (from both
sides) at either 3 or 1.
(a) Find the largest interval contain x = 2 on which f is continuous. Explain.
(b) Find the largest interval contain x = 5 on which f is continuous. Explain.
(c) Find the largest interval contain x = 7 on which f is continuous. Explain.
(d) Find the largest interval contain x = 9 on which f is continuous. Explain. Answers to you try it 9. 7 : (a) ( 1, 5 )
since f is not left-continuous at 5 nor
right-continuous at 1. (b) [ 5, 9 ). (c)
[ 5, 9 ). (d) [ 9, 11 ].
EXAMPLE 9. 10 (Intervals of Continuity). Determine the largest intervals of continuity for
f ( x ) =
8
<
:
x
x 1
if x > 2
2 if x = 2
x
2 x
x
5 x + 6
if x < 2
.
SOLUTION. The piecewise function consists of two rational functions and a value
at a single point. On ( 2, • ) , f ( x ) =
x
x 1
is continuous by Theorem 8. 1 since the
denominator is never 0. So f is continuous at least on ( 2, • ). Is f right-continuous at
2? Well, we know that f ( 2 ) = 2. And
lim
x! 2
f ( x ) = lim
x! 2
x
2
x 1
=
5
1
= 5 6 = 2.
So f is not right continuous at 2. So one of the intervals of continuity is ( 2, • ).
On the interval ( • , 2 ) , f ( x ) =
x
2 x
x
5 x + 6
=
x ( x 2 )
( x 2 )( x 3 )
is rational and continuous by
Theorem 8. 1 since the denominator is never 0 there. Is f left-continuous at 2?
lim
x! 2
f ( x ) = lim
x! 2
x
2
2 x
x
2
5 x + 6
= lim
x! 2
x
x 3
=
2
1
= 2 = f ( 2 ).
So f is left-continuous at 2. Therefore, f is continuous on ( • , 2 ].
Note, however that f is not continuous at 2 since the two one-sided limits are
different there.
EXAMPLE 9. 11 (Intervals of Continuity). Determine the largest intervals of continuity for
f ( x ) =
(
x + 1
x
if x > 0
x
2
x 1 if x 0
.
math 130 continuity: days 8 and 9 12
SOLUTION. The piecewise function consists of a rational function and a polynomial.
On ( 0, • ) , f ( x ) =
x + 1
x
is continuous by Theorem 8. 1 since the denominator is never
- So f is continuous at least on ( 0, • ). Is f right-continuous at 0? Well, we know that
f ( 0 ) = 0
2
0 = 0. And
lim
x! 0
f ( x ) = lim
x! 0
1
z }| {
x + 1
x
|{z}
0
= + •
So f is not right continuous at 0. So one of the intervals of continuity is ( 0, • ).
On the interval ( • , 0 ) , f ( x ) = x
2
x 1 is polynomial and is continuous by
Theorem 8. 1. Is f left-continuous at 0?
lim
x! 0
f ( x ) = lim
x! 0
x
2
x 1
Poly
= 1 = f ( 0 ).
So f is left-continuous at 0. Therefore, f is continuous on ( • , 0 ]. Note, however that
f is not continuous at 0 since the two one-sided limits are different there.
EXAMPLE 9. 12 (Intervals of Continuity). Determine the largest intervals of continuity for
f ( x ) =
(
x + 2 if x 3
x
2
x 1 if x < 3
.
SOLUTION. The piecewise function consists of two polynomials. On ( 3, • ) , f ( x ) =
x + 2 is a continuous polynomial. Is f right-continuous at 3? Well, we know that
f ( 3 ) = 3 + 2 = 5. And
lim
x! 3
f ( x ) = lim
x! 3
x + 2 = 5
So f is right continuous at 3. So one of the intervals of continuity is [ 3, • ).
On the interval ( • , 3 ) , f ( x ) = x
2
x 1 is polynomial and is continuous by
Theorem 8. 1. Is f left-continuous at 0?
lim
x! 3
f ( x ) = lim
x! 3
x
2
x 1
Poly
= 5 = f ( 3 ).
So f is left-continuous at 3. Therefore, f is continuous on ( • , 0 ]. In fact, because the
two one-sided limits at 3 are equal,
lim
x! 3
f ( x ) = lim
x! 3
f ( x ) = 5,
we have lim
x! 3
f ( x ) = 5 = f ( 3 ). So f is continuous at 3 (and everywhere else since it
is a polynomial if x 6 = 3. So f ( x ) is actually continuous for all x, i.e., on ( • , • ).
