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1.6 Continuity
Definition 8 (continuity):
Let f x  be defined for every x in an interval containing the number
x0 . Then, f is continuous at x0 if
lim f x   f x0  .
xx0
This means that the following 3 conditions hold:
(i) f  x0  exists (i.e., x0 is in the domain of f ).
f x  exists.
(ii) xlim
x
0
f x   f x0  .
(iii) xlim
x
0
Note:
A function f is continuous over (or in) the open interval
or the closed interval
a, b
a, b
if f is continuous at every point in
that interval.
Example 31:
Let
f  x   x 2 . Then, for any real number x0 ,
lim f x   lim x 2  x02  f x0  .
xx0
So, f
Note:
xx0
is continuous at every real number.
let
f  x   c0  c1 x  c2 x 2    cn x n
polynomial. By theorem 2,
lim f x   f x0  .
xx0
1
be
a
Therefore, every polynomial is continuous at every real number.
Similarly, let
c0  c1 x  c2 x 2    cn x n
p x 
f x  

q x  d 0  d1 x  d 2 x 2    d m x m
be a rational function. If qx   0 , then
px  px 
px  xlim
 x0
0
lim f  x   lim


 f x0  .
x x0
x x0 q  x 
lim qx  qx0 
x x0
Therefore, any rational function is continuous at all points x0 at
which the denominator, qx0  , is nonzero.
◆
Definition 9 (discontinuity):
If the function f x  is not continuous at x0 , then f is said to be
discontinuous at x0 if one (or more) of the 3 conditions given in the
definition for continuity fail to hold. That is, f is discontinuous at x0
under any of the following conditions:
(i) f  x0  does not exist.
f x  does not exist.
(ii) xlim
x
0
f x  exists but is not equal to f x0  .
(iii) xlim
x
0
Example 8 (continue):
Let
f x  
x is not continuous for x  0 since this function is not defined
for negative values.
2
Example 4 (continue):
Let
 x  1, x  2
f x   
.
2 x  1, x  2
This function is discontinuous at
x2
since
lim f x 
x 2
does not exist.
Note: in example 4, the left-hand and right-hand limits of f x  at
x  2 exist but are not equal. In this situation, f is said to have a
jump discontinuity at x  2 .
Theorem 6:
Let the function f and g be continuous at
x0 ,
constant. Then the following functions are continuous at
and let c be a
x0 :
(i) cf (ii) f  g (iii) f  g (iv) f / g if g x0   0 .
Theorem 7 (Continuity theorem):
If
f
is
continuous
at
a
and
if
lim f x   a ,
xx0
then
lim f g x   f a . . That is,
xx0
lim f g x   f  lim g x . .
x x0
 xx0

Example 32:
Let
Let
lim g x   c  0 . Compute lim
xx0
x x0


g x   g 4 x  .
f  x   x  x 4 . Then, f g x   g x   g 4 x. Since f x 
is continuous at
x  c , then by theorem 7,
3
lim f g x   f  lim g x   f c   c  c 4 .
x x0
 xx0

Theorem 8:
Let the function
differentiable at
f
be differentiable at
x0
. Then,
f
is
x0 .
[Intuition:]
 f x   f x0 x  x0 
x x
x  x0 
f x   f x0 
 lim
lim x  x0 
x x
x x
lim  f x   f x0   lim
x x0
0
0
x  x0
0
 f ' x0   0
0
◆
Note:
A function that is continuous at a point is not necessarily
differentiable at that point. For example,
x
is continuous at 0 but
not differentiable at 0.
Theorem 9:
If the function f is continuous on the closed, bounded interval
a, b ,
then f is bounded above and below in that interval. That is, there exist
numbers m and M
a, b . Moreover, if
such that m  f x  M for every x in
m is the greatest lower bound for f on a, b
and M is the least upper bound for f on
4
a, b , then there exist
numbers
x1
and
x2
in
a, b
f x1   m and
such that
f x2   M .
Theorem 10 (Intermediate-value theorem:
Let the function f
be continuous on the closed, bounded interval
a, b . Then, if
c is any number between f a  and f b , there is
a number x in
a, b
Example 33:
Using intermediate-value
f x   x 3  x 2  x  1
such that f x   c .
theorem,
justify
in the interval
The justifications are as follows. Since
that
root
of
is
a
f 0  03  02  0  1  1
and
0,1 .
f 1  13  12  1  1  2 , there must be a number x
 1  f x   0  2 .
5
there
in
(0,1)
such that