1.3 Continuity and Its Consequences
1. Definition:
Graphically (visually), a function f is continuous at x ! a if the graph of f has no break up (hole, vertical
asymptote, jump) at x ! a.
Definition A function f is continuous at x ! a when
!i" f!a" is defined;
!ii" lim x"a f!x" exists; and
!iii" lim x"a f!x" ! f!a".
Otherwise, f is said to be discontinuous at x ! a.
Note that f is continuous at x ! a if and only if all these 3 conditions are satisfied. So, if one of the
conditions does not hold, we can conclude that f is not continuous at x ! a.
Use the definition of continuity to explain the break up of the graph of f at x ! a:
! hole - f!a" is not defined or lim x"a f!x" ! f!a"
! vertical asymptote - f!a" is not defined
! jump - lim x"a f!x" does not exist.
Example Page 112 #10
Example Determine if f is continuous at x ! 2.
2
a. f!x" ! x " 4
x"2
b. f!x" !
x2 " 4
x"2
4
if x ! 2
if x ! 2
a. f is not continuous at x ! 2 because f!x" is not defined at x ! 2.
b. Know f!2" ! 4. Now check to see if lim x"2 f!x" is 4. Check:
2
lim f!x" ! lim x " 4
both numerator and denominator are 0 at x ! 2
x"2
x"2
x"2
!x " 2"!x # 2"
! lim
! lim !x # 2" ! 4
x"2
x"2
x"2
Hence, f is continuous at x ! 2 by definition.
Example Let f!x" !
x2
if x $ 2
3
if x ! 2 .
3x " 2
if x % 2
a. Evaluate lim x"2 f!x".
b. Is f!2" defined? If so, what is f!2".
c. Determine if f is continuous at x ! 2.
a. lim x"2 " f!x" ! lim x"2 " x 2 ! !2" 2 ! 4,
b. f!2" is defined and f!2" ! 3
1
lim x"2 # f!x" ! lim x"2 # !3x " 2" ! 4, so, lim x"2 f!x" ! 4
c. f is not continuous at x ! 2 because lim x"2 f!x" ! f!2".
Example Find all discontinuous points of f!x".
2
a. f!x" ! 2x # 1
x #x"6
b. f!x" ! x csc x
c. f!x" !
2x
if x $ 0
sin x
if x ! 0
x " ! if x % 0
a. f!x" is not defined when x 2 # x " 6 ! !x # 3"!x " 2" ! 0. So, f is not continuous at x ! "3, x ! 2
b. f!x" ! x csc x ! x , f!x" is not defined when sin x ! 0, that is, x ! n!, n ! 0, &1, &2, ' . So,
sin x
f!x" is not continuous at x ! n!, n ! 0, &1, &2, ' .
c. f is defined everywhere. f is continuous for all x $ 0 and f is continuous for all x % 0. Check if f is
continuous at x ! 0. Know f!0" ! sin!0" ! 0.
lim" f!x" ! lim" 2x ! 0, lim# f!x" ! lim# !x " !" ! "!, so lim f!x" DNE
x"0
x"0
x"0
x"0
x"0
So, f is not continuous at x ! 0.
2. Continuity of Special Functions:
These functions are continuous everywhere in their domains.
a. All polynomials are continuous everywhere.
b. Functions e x , sin x, cos x are continuous everywhere.
c. Function ln x continuous everywhere for x % 0.
3. Continuity on a Closed Interval:
Definition:
If f is continuous at every point on an open interval a, b , f is said to be continuous on a, b . f is said
to be continuous on the closed interval a, b if f is continuous on a, b and
lim f!x" ! f!a"
and lim" f!x" ! f!b".
x"a #
x"b
If f is continuous at every point on "#, # , f is said to be continuous (everywhere).
Example Determine the intervals on which f!x" is continuous.
a. f!x" !
x2 " 2
a. Find the domain of f!x" : D f !
x2 " 2 $ 0
#
b. f!x" ! !2x # 1" 5/2
x | x2 " 2 $ 0
x2 $ 2
#
x|x$
!
x $
c. f!x" ! ln!x 2 # x " 2"
2 or x %
2 #
x$
2
2 or
x%" 2
We know f is continuous on every point in "#, " 2 &
2 , # . Check if f is continuous at
x ! " 2 and x ! 2 . By definition of continuity of a closed interval:
lim "
x"" 2
x2 " 2 ! 0 ! f " 2
and
lim #
x" 2
x2 " 2 ! 0 ! f
Hence, f is continuous on !"#, " 2 $ & % 2 , #".
b. Note that f!x" ! !2x # 1" 5/2 ! !2x # 1" 5 . Find the domain of f!x" :
D f ! x | 2x # 1 $ 0 ! x | x $ " 1
2
2
.
2 .
f is continuous on " 1 , # . Check if f is continuous at x ! " 1 :
2
2
5
lim1 # f!x" ! lim1 # !2x # 1" ! 0
x"" 2
x"" 2
Hence, f is continuous on %" 12 , #".
c. Find the domain of f!x" : D f ! x | x 2 # x " 2 % 0
x 2 # x " 2 ! !x # 2"!x " 1" % 0
Ways to solve the inequalities: !x # 2"!x " 1" % 0
i. Graphically, the graph of !x # 2"!x " 1" is as follows.
20
15
10
5
-4
-2
0
2 x
4
y ! !x # 2"!x " 1"
!x # 2"!x " 1" % 0 # x $ "2, or x % 1.
ii. Algebraically, let g!x" ! !x # 2"!x " 1". Check
g!"3" ! !"1"!"4" ! 4 % 0, g!0" ! !2"!"1" ! "2, g!2" ! !3"!1" ! 3 % 0
So, !x # 2"!x " 1" % 0 # x $ "2, or x % 1.
Hence, D f ! "#, " 2 & 1, # . Hence, f is continuous on "#, " 2 & 1, # .
4. Rules: Suppose that f and g are continuous at x ! a. Then
a. cf is continuous at x ! a
b. f & g is continuous at x ! a
c. fg is continuous at x ! a
d. gf is continuous at x ! a if g!a" ! 0 and is discontinuous at x ! a if g!a" ! 0
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