GUC

MATH103
For Engineering
Winter 2023

Lecture # 8
Chapter 2
App’s of limits: Continuity of functions
1
Continuity at a point

2
 Computational example
Example:
−x
Discuss the continuity of the function f ( x) = e at the point x = 0.
Solution: The given function can be redefined as follows:
−x e x , if x  0
f ( x) = e =  −x
e , if x  0

At x = 0, we test the three conditions of continuity of a

function at a point as follows:
1) f (0) = e0 = 1, by the definition of the function.
2) We study the limit of the function as x approaches 0 as
follows:
lim f ( x ) = lim e x = 1, lim f ( x ) = lim e − x = 1  lim f ( x ) = 1
x →0 − x →0− x →0+ x →0+ x →0

3) Since lim f ( x ) = 1 =
x →0
f (0),

it follows that f (x) is continuous at x = 0. 3

 Relation between limits and continuity
A function is not continuous at a point x = a, if one of the following happens:
1- f(a) is not defined.
Or
2- f (a) is defined. BUT, lim f (x) = DNE (as x→a).
Or
3- f (a) is defined AND lim f (x) = exists (as x→a) BUT they are Not equal.

Graphical Example:
Discuss all types of limits
for the shown function at
the points 0, 1, 2, 3, and 4.

No limit from the left

Limit from the left is

Limit from the left equals 4
different from limit from
to limit from the right No limit from the right
the right
 Types of discontinuities
(1) Removable discontinuity lim f ( x ) = Exists
x →a
A function f is said to have a removable discontinuity at a point “x = a”, If
limit of f exists as x tends to “a” such that
either f is not defined at x = a ( or defined differently).

sin x x + 1 if x  0
f ( x) = f ( x) = 
x 2 if x = 0

Removable
Discontinuities f( 0 ) = 2 Defined
f( 0 ) is not defined ,
at x = 0 and lim f ( x ) = 1
sin x x →0
and lim =1
x→0 x
5
 (2) Jump discontinuity lim f ( x )  lim f ( x )
x →a + x →a −

A function has a jump discontinuity at a point “x = a”, If both (the left limit and
the right limit) “at x = a” exist but they are not equal.

0, x  0
f ( x) = 
1, x  0

lim− f ( x ) = 0 lim+ f ( x ) = 1
x →0 x →0
f(x) has a jump
discontinuity at x =0
| x|
Exercise: Discuss the continuty of : f ( x ) = at x = 0. 6
x
 (3) Infinite discontinuity lim f ( x ) = 
x →a 
A function has an infinite discontinuity at a point x = a, If at least (one of the left
or right sided limits or both) at “x = a” equals infinity.

1
g( x ) =
( x + 3) 2

g(x) has an infinite

discontinuity at x = -3

1 1
lim− ( x + 3)2 = lim+ ( x + 3)2 = 
x → −3 x → −3

1
Exercise: Discuss the continuty of : f ( x ) = at x = −5.
x+5 7
 (3) Infinite discontinuity, (another example)
1, x  0

f ( x) =  1
 x , x  0

lim− f ( x ) = 1
x →0

lim+ f ( x ) = 
x →0

f(x) has an infinite

discontinuity at x =0 8
Summary of the types of discontinuities

9
 Continuity on a domain
A function is continuous on a closed interval [a, b]
if is continuous at all interior points as well as
continuous at the two-end points a, b.

left end point interior point right end point

10
 Example: Discuss continuity of the shown
function on the closed interval [0, 4]

jump discontinuity
at x = 1 removable discontinuity at x = 2, 4 11
 Basic properties of continuous functions

Any Polynomial is continuous on R “the whole real line” 12

Example: Discuss the continuity of each of the following functions on
its domain:  x 2 − x − 12
 if x  −3 −x
(i ) f ( x ) =  x+3 ( ii ) f ( x ) = e ( HW )
− 5 if x = −3

Solution: x 2 − x − 12
(i) On ]-infinity, -3[ & On ]-3, infinity[, is a x+3
quotient of two polynomials, so it is continuous.

(ii)@ x = -3, We examine the three conditions of continuity at a point:

1) f (-3) = -5, defined.
2) Limit: x − x − 12
2
( x − 4)( x + 3)
lim f ( x ) = lim x + 3 = lim x + 3 = lim ( x − 4) = −7
x → −3 x → −3 x → −3 x → −3

3)  lim f ( x ) = −7 
x → −3
f ( −3) = −5

 f(x) is NOT continuous at x = −3 and has a removable discontinu ity.

From (i), (ii), we say that f is continuous on R-{-3} 13

 Computational exercise
For what value of the constant c is the function
f(x) continuous for all values of x ?
cx + 1, if x  3
f ( x) =  2
cx − 1, if x  3
Solution : Since f is continuous on R
 f is continuous at x = 3 c = 1/3
 lim f ( x ) = Exists
x→3

 lim− f ( x ) = lim+ f ( x )  lim− (cx + 1) = lim+ (cx 2 − 1)

x→3 x→3 x→3 x→3

3c + 1 = 9c − 1  6c = 2  c = 1 / 3
14
 Challenging problem

Discuss continuty of f(x) for all x

x sin x
f ( x) =
x2 + 2

This function is
continuous for all
real values “x”

15
The intermediate value theorem “I.V.T.”

Notice that we can have

more than one value c

16
 Importance of the continuity of the function

The opposite figure

shows that if the function
is discontinuous at at
least one point then the
conclusion of the theorem
may be false.

17
Important Application: “corollary of the I.V.T.” If f (x) is:
1) continuous on [a, b], and
2) f (a) and f (b) have different signs.
Then, there is at least one-point c in ] a, b [ for which f (c) = 0.
Example: Show that the
equation: x3 – x – 1 = 0 has
at least one solution
between x = 1 and x = 2.
Solution:
Let f (x) = x3 – x – 1.
Then, it is clear that:
f is continuous on the
interval [1, 2],
and
f (1) = -1 < 0 & f (2) = 5 > 0.
Then, by using the above
corollary: the equation
x3 – x – 1 = 0 has at least
one solution between: 18
x = 1 and x = 2.
 Practical problem
Use the intermediate value theorem to show that
among all circles with radius no larger than 10 cm.,
there is one whose area is 200cm2.
Solution:
The area A(r) of a circle with radius r is given by A(r) = πr2.
Since A(r) is continuous function on the interval [0, 10]
and for r = 0, A(r) = 0, and for r = 10, A(r) = 100π = 314
and 200 lies between 0 and 314, then by the intermediate
value theorem there is an r between o and 10 such that
A(r) = 200.

19
Thank you

20