2.
6 Continuity of a Function at a Number
ILLUSTRATIONS : Consider the following functions; 1. f ( x ) = x 2 x + 6
2
Since f is a polynomial function then it is continuous at every real number.
ILLUSTRATIONS : Consider the following functions;
x+ 1 2. g ( x ) = x 1 Since g is a rational function then it is continuous at every real number except at x = 1 .
EXAMPLE 1 : Test the given function for continuity at 0;
x+ 3 , x 0 f ( x) = 4 , x< 0
Test for continuity at x = 0. i. f ( 0 ) = 3 ii. lim+ f ( x ) = lim+ ( x + 3 ) = 3
x 0 x 0 x 0
lim f ( x ) = lim 4 = 4 +
x 0
lim f ( x ) does not exist. x 0
iii. lim f ( x ) f ( 0 )
x 0
The function f has an essential discontinuity at x = 0.
Geometrically, a function is continuous at a number if its graph has NO BREAK at that number.
EXAMPLE 2: Test the given function for continuity at 0;
x + 1 , x 0 g( x ) = , x= 0 0
2
Test for continuity at x = 0. i. g ( 0 ) = 0 ii. lim g ( x ) = lim ( x + 1) = 1 x 0
2 x 0
x2 + 1 , x 0 g( x ) = , x= 0 0
iii. lim g ( x ) g ( 0 )
x 0
The function g has a removable discontinuity at x = 0.
ORIGINAL FUNCTION
x + 1 , x 0 g( x ) = , x= 0 0
2
REDEFINED FUNCTION
x + 1 , x 0 G( x ) = , x= 0 1
2
g is a rational function, g is continuous at each number in its domain. Dg = { a / a 3,3}
Since g(-3) and g(3) are undefined, g is discontinuous at -3 and 3.
a 81 g ( a) = 2 a 9
4
a 81 EXAMPLE 3: g ( a ) = a 2 9 .
4
WHAT TYPE OF DISCONTINUITY ?
( a 3) ( a + 3) ( a 2 + 9 ) ( a 3) ( a + 3)
= a 2 + 9, a 3,3
a 4 81 g ( a) = 2 a 9
at a = -3: g(-3) is undefined.
a 4 81 . lim g ( a ) = lim 2 a 3 a 3 a 9
= lim
a 3
a 3) ( a + 3) ( a 2 + 9 ) (
( a 3) ( a + 3)
= 18
(The limit exists.)
Hence, g has a removable discontinuity at a = -3.
a 4 81 g ( a) = 2 a 9
at a = 3: g(3) does not exist. .
a 4 81 lim g ( a ) = lim 2 a 3 a 3 a 9
= lim
a 3
a 3) ( a + 3) ( a 2 + 9 ) (
( a 3) ( a + 3)
= 18
(The limit exists.)
Hence, g has a removable discontinuity at a = 3.
Since lim g ( a ) and lim g ( a ) exist, a 3 a 3 the discontinuities at -3 and 3 are removable. Redefining g by letting g(-3) = 18 and g(3) = 18 makes the new function a 4 81 , if a 3 2 g ( a) = a 9 18, if a = 3 continuous at every number.
a 4 81 , if a 3 2 g ( a) = a 9 18, if a = 3
EXAMPLE 4:
x, x 1 h ( x) = x, x < 1
The only possible discontinuity of h is at x = 1.
h ( 1) = 1
x 1
lim h ( x ) = lim ( x ) = 1
x 1
lim h ( x ) = ?
x 1
lim h ( x ) does not exist
x 1
x 1
lim h ( x ) = lim x = 1 + +
x 1
Thus, h is essentially discontinuous at x = 1.
x, x 1 h ( x) = x, x < 1
Note that the break in the graph of f cannot be bridged by the addition or replacement of a single point.
( x + 4) 2 , x< 2 EXAMPLE 5: h ( x ) = 4 x2 , 2 x < 2 x 2 , x> 2
Possible points of discontinuity: x = -2 or x = 2 at x = -2:
x 2
lim h ( x ) = ?
h ( 2) = 0
lim h ( x ) = lim ( x + 4 ) = 4
2 2 x 2
x 2
lim + h ( x ) = lim + 4 x = 0
2
x 2
lim h ( x ) does not exist
x 2
Thus, h has an essential discontinuity at x = -2.
at x = 2:
( x + 4) 2 , x < 2 h ( x ) = 4 x2 , 2 x < 2 x 2 , x> 2
lim h ( x ) = ?
x 2
h ( 2) = ?
It does not exist.
x 2
lim h ( x ) = lim 4 x 2 = 0
x 2
x 2
lim h ( x ) = lim x 2 = lim+ ( x 2 ) = 0 + +
x 2 x 2
lim h ( x ) = 0 (The limit exists.) x 2
Thus, h has a removable discontinuity at x = 2.
( x + 4) 2 , x< 2 2 h ( x) = 4 x , 2 x < 2 x 2 , x> 2
h can be redefined such that it will be
continuous at x = 2.
( x + 4) 2 , x< 2 h ( x ) = 4 x2 , 2 x < 2 x 2 , x 2
Challenge Question Let
2 x 2 + 1, x 1 . Find the f ( x ) = ax + b, 1 < x < 1 x3 , x 1
values of a and b so that f is continuous everywhere. Solution: From the definition of f, it is clear that f is continuous at all real numbers except possibly at 1 and 1.
at x = -1:
lim f ( x ) = ?
f ( 1) = 3
x 1
x 1
lim f ( x ) = lim 2 x 2 + 1 = 3
x 1
2 x 2 + 1, x 1 f ( x ) = ax + b, 1 < x < 1 x3 , x 1
x 1
lim+ f ( x ) = lim+ ( ax + b ) = a + b
x 1
For the limit to exist and the function to be continuous at x = -1, a + b = 3 at x = 1: f ( 1) = 13 = 1
x 1 x 1
x 1
lim f ( x ) = lim ( ax + b ) = a + b
x 1
lim f ( x ) = ?
lim+ f ( x ) = lim x 3 = 1 +
x 1
For the limit to exist and the function to be a+ b= 1 continuous at x = 1,
Thus, for f to be continuous at all real numbers including 1 and 1, a and b should satisfy the following equations:
a+ b= 1
and a + b = 3 Solving the system .
a+ b= 3 a+ b= 1
we get a = 1 and b = 2 .
x 3
lim f ( x) lim+ f ( x) lim f ( x)
x 0
x 2
lim f ( x ) lim f ( x ) +
4 3 1 -3 -1 1 3
x 3
x 2
x 2
x 3
lim f ( x)
lim f ( x) lim f ( x)
x 1
x 3
lim f ( x) + lim f ( x)
x +
x = -2
x=2
lim f ( x)
Pts. of discontinuity: -3 removable
-2
essential
