l6 continuity

LIMITSOF

FUNCTIONS

CONTINUITY

Definition 1.5.1 (p. 110)

If one or more of the above conditions fails to hold at C the function is said to be discontinuous.

DEFINITION: CONTINUITY OF A FUNCTION

Theorem 1.5.3 (p. 113)

Question 8

EXAMPLE

Solution:

( )( ) ( )2 2 36

3

x xx xf x

x

+ −− −= =− 3x −

( )

2

2 3

x

f x x where x

= +

∴ = − ≠

1. Given the function f defined as ,draw a sketch of the graph of f, then by observing where there are breaks in the graph, determine the values of the independent variable at which the function is discontinuous and why each is discontinuous.

( )2 6

3

x xf x

x

− −=−

•

••

•

y

x

Test for continuity: at x=3

1.f(3) is not defined; since the first condition is not satisfied then f is discontinuous at x=3.

Question 8

2. Given the function f defined as

draw a sketch of the graph of f, then by observing where there are breaks in the graph, determine the values of the independent variable at which the function is discontinuous and why each is discontinuous.

EXAMPLE

( ) if

if

2 63

32 3

x xx

f x xx

− − ≠= − =

•

••

•

y

x

•

Solution:

( )

( ) ( )

( )

3x at ousdiscontinu is f therefore

)3(f)x(flim but exists itlim The

2)3(f and 5)x(flim .3

523

2xlim 3x

2x3xlim

3x

6xxlim)x(flim .2

2 to equal is and defined is 3f .1

:continuity for Test

3x

3x

3x

3x

2

3x3x

=

≠

===+=

+=−

+−=

−−−=

→

→

→

→

→→

Question 8Question 8

EXAMPLE

3. Given the function f defined as ,

draw a sketch of the graph of f, then by observing where there are breaks in the graph, determine the values of the independent variable at which the function is discontinuous and why each is discontinuous.

( ) if

if

2

10

2 0

xf x x

x

≠= =

20

20

1 1lim

0

1 1lim

0

0

x

x

x

x

x is a VA

+

−

→+

→+

= = +∞

= = +∞

∴ =

VA a is 0x

0x if x

1)x(f

:Graph

2

=

≠=

2

2

1lim 0

1lim 0

0

x

x

x

x

y is a HA

→+∞

→−∞

=

=

∴ =

HA

0x at ousdiscontinu

is f then satisfied not is

condition ondsec the Since

exists not does lim .2

defined;2)0(f .1

:continuity for Test

0x

=

+∞==

→

Question 8Question 8

Solution:y

x

1

-1 1

•

Figure 1.5.1 (p. 110)

The figure above illustrates the function not defined at x=c, which violates the first condition.

The figure above illustrates that the limit coming from the right and left both existbut are not equal, thus the two sided limit does not exist which violates the second condition. This kind of discontinuity is calledjump discontinuity.

Figure 1.51 (p. 110)

The figure above illustrates that the limit coming from the right and left of c are both , thus the two sided limit does not exist which violates the second condition. This kind of discontinuity is called infinite discontinuity.

∞+

The figure above illustrates the function defined at c and that the limit coming from the right and left of c both exist thus the two sided limit exist. But which violates the third condition. This kind of discontinuity is calledremovable discontinuity.

)x(flim)c(fcx→

≠

2

y

x

)x(fy =

2x

4x)x(f given

; 2x at continuous is function the whether eminDeter2

−−=

=

( ) ( )( )

2x at continuousdis hence and

2x at undefined is )x(f thus

2x but ;422)2(f

2x)x(f

2x

2x2x)x(f

==

≠=+=

+=−

−+=

4

2

4

y

x

)x(gy =

3 •

=

≠−−

=

=

2x ,3

2x ,2x

4x)x(g given

; 2x at continuous is function the whether eminDeter2

( ) ( )( )

2x at continuousdis hence

)2(g)x(glim the cesin

3)2(g

4)x(glim thus

4)x(glim)x(glim

2x)x(g

2x

2x2x)x(g

2x

2x

2x2x

=

≠=

=

==+=

−−+=

→

→

→→ −+

Removable Discontinuity

=

≠−−

=

=

2x ,4

2x ,2x

4x)x(h given

; 2x at continuous is function the whether eminDeter2

( )( )( )

2x at continuous hence

)2(h)x(hlim the cesin

4)2(h

4)x(hlim thus

4)x(hlim)x(hlim

2x)x(h

2x

2x2x)x(h

2x

2x

2x2x

=

==

=

==+=

−−+=

→

→

→→ −+

2

4

y

x

•