Continuity

Section 2.5

Prepared by:
Prof. Wafik Lotfallah
Prof. Maged G. Iskander
Dr. Kamal Tawfik
Lecture 7 Objectives
 For a function f given by its graph or its defining
expression:
 State the numbers at which f is discontinuous.
 State the intervals on which f is continuous.
 Decide if a certain discontinuity is removable.
 Check the continuity of a function f given by a
case-definition.
 Apply the Intermediate Value Theorem to show
the existence of a solution to an equation.
Continuity at a point

Note: The above definition

requires that:
 𝑓(𝑎) is defined.
 The limit from the left exists.
 The limit from the right exists.
 All of these are equal.
Geometrically, you can think of a function
that is continuous at every number in an
interval as a function whose graph has no
break in it.

 The graph can be drawn without removing

your pen from the paper.
Example 1
The figure shows the graph of a function f. At which
numbers is f discontinuous? Why?
At 𝒙 = 𝟏.
The function is discontinuous, as the graph has a break
there.
The official reason that 𝑓 is discontinuous at 1 is that 𝑓(1) is
not defined.

At 𝒙 = 𝟑.
The graph also has a break. However, the reason for the
discontinuity is different.
Here, 𝑓(3) is defined,
but lim 𝑓(𝑥) does not exist as
x→3
(the left and right limits are different.)
So, 𝑓 is discontinuous at 3.
What about 𝒙 = 𝟓?
Here, 𝑓(5) is defined and lim𝑓(𝑥) exists (as the left and
x→5
right limits are the same).

However, 𝑓 is discontinuous at 𝒙 = 𝟓 since

lim f ( x)  f (5)
x →5
Theorem

Example 𝟐
𝑥 3 +2𝑥 2 −1
Find lim
5−3𝑥
𝑥→−2
Solution.
𝑥 3 +2𝑥 2 −1
The function 𝑓(𝑥) =
5−3𝑥
is rational, so by Theorem 5 it is continuous on its
domain, which is 𝑥 𝑥 ≠ 53}.
Therefore,

Example 𝟑
Where is the function 𝑓 below continuous?
𝑥2 − 1
𝑓(𝑥) = 3
𝑥 − 4𝑥 2 − 5𝑥
Solution
We can write the function 𝑓 as
𝑥2 − 1 𝑥2 − 1
𝑓 𝑥 = 2
= .
𝑥(𝑥 − 4𝑥 − 5) 𝑥(𝑥 − 5)(𝑥 + 1)
Since the function 𝑓 is rational, so by Theorem 5 it is
continuous on its domain, which is 𝑥 𝑥 ≠ −1, 0, 5}.

Remark:
Although 𝑓(𝑥) is not defined at 𝑥 = −1, its limit exists. We
can calculate its limit as
𝑥2 − 1
lim 𝑓 𝑥 = lim
x→−1 x→−1 𝑥(𝑥 − 5)(𝑥 + 1)
𝑥 − 1 (𝑥 + 1) 2
= lim =−
x→−1 𝑥(𝑥 − 5)(𝑥 + 1) 6

In this case we say that:

𝑓(𝑥) has a removable discontinuity at 𝑥 = 1.
 Types of Discontinuity at (𝒙 = 𝒄):
 Removable Discontinuity: When the limit exists at 𝒙 = 𝒄,
even if 𝒇(𝒄) is undefined.

 Jump Discontinuity:
When the two-side limit exist, but
not equal at 𝒙 = 𝒄

Infinite Discontinuity: When, at least, one of the one-sided

limits of the function is infinite.
E.g., the following example at x = 0.
One Sided Continuity
Example 4
For the function f, state the numbers at which f is
continuous from the right/left?
Continuity on Intervals
Theorem
In general:
The following functions are continuous on their domains:
 Polynomials
 Rational Functions
 Root Functions
 Trigonometric Functions
 Exponential Functions
 Logarithmic Functions
Example 𝟓
Where is the function f below continuous?
1
f ( x) = sin + tan x
x
Answer:

 This happens when x is an odd integer multiple of  /2, so

y = tan θ has infinite discontinuities when
θ =  /2, 3 /2, 5 /2, and so on. Then the function is
continuous for x > 0 and x ≠ (/2)2 , (3/2)2, (5/2)2, …..
Example 6
Find the values of the constants a and b, that make the
function f below continuous.

 3 if x0

f ( x) = ax + b if 0  x 1
1 + x 2 if x 1

Illustrate the continuous function with a sketch.
Solution
Clearly, each branch of 𝑓(𝑥) is continuous on its domain,
also we must have continuity at the boundary points.
At 𝒙 = 𝟎:  3 if x0

lim− 𝑓 𝑥 = lim− 3 = 3. f ( x) = ax + b if 0  x  1
𝑥→0 𝑥→0 1 + x 2
 if x 1

lim+ 𝑓 𝑥 = lim+ 𝑎𝑥 + 𝑏 = 𝑏.
𝑥→0 𝑥→0
We must have
𝑓 0 = lim− 𝑓 𝑥 = lim+ 𝑓 𝑥
𝑥→0 𝑥→0
which gives 𝑏 = 3.
At 𝒙 = 𝟏: x0
 3 if
lim− 𝑓 𝑥 = lim− 𝑎𝑥 + 3 = 𝑎 + 3. 
𝑥→1 𝑥→1 f ( x) = ax + b if 0  x 1
1 + x 2 if x 1

lim+ 𝑓 𝑥 = lim+ 1 + 𝑥 2 = 1 + 1 = 2.
𝑥→1 𝑥→1
We must have
𝑓 1 = lim− 𝑓 𝑥 = lim+ 𝑓 𝑥
𝑥→1 𝑥→1
⇒ 𝑎 + 3 = 2 ⇒ 𝑎 = −1.
Now, we can rewrite the function
𝑓(𝑥) as:
3 𝑖𝑓 𝑥 < 0
𝑓 𝑥 = ൞−𝑥 + 3 𝑖𝑓 0 ≤ 𝑥 ≤ 1
1 + 𝑥 2 𝑖𝑓 𝑥>1
The Intermediate Value Theorem

Note that:
the value N can be taken on once [as in part (a)] or more
than once [as in part (b)].
Application of the theorem
Example 𝟕
Show that there is a root (solution) of the equation:
4𝑥 3 − 6𝑥 2 + 3𝑥 − 2 = 0
between 1 and 2.
Solution
Let 𝑓(𝑥) = 4𝑥 3 − 6𝑥 2 + 3𝑥 − 2.
We are looking for a solution of the given equation, that is, a
number 𝑐 between 1 and 2 such that 𝑓(𝑐) = 0.
We have
𝑓 1 = 4 − 6 + 3 − 2 = −1 < 0
and 𝑓 2 = 4(2)3 −6 2 2 + 3(2) − 2 = 12 > 0
Thus
𝑓(1) < 0 < 𝑓(2).
Now 𝒇 is continuous since it is a polynomial, so the IVT
says there is a number 𝑐 between 1 and 2 such that 𝑓(𝑐) = 0.
In other words, the equation
4𝑥 3 − 6𝑥 2 + 3𝑥 − 2 = 0
has at least one root 𝑐 in the interval (1, 2).

Example 8
Show that there is a solution to the equation:
ln 𝑥 = 3 − 2𝑥 .
Solution
Let 𝑓 𝑥 = 3 − 2𝑥 − ln 𝑥 .
Since 𝑓(𝑥) is continuous on the interval [1,2] and we have

𝑓 1 =3−2−0=1 >0
and 𝑓 2 = 3 − 2 2 − ln 2 = −1 − ln 2 ≈ −1.7 < 0.

Since 𝑓 2 <0<𝑓 1 ,
by IVT, there is a number 𝑐 in (1, 2) such that 𝑓 𝑐 = 0.
Thus, there is a root of the equation 3 − 2𝑥 − ln 𝑥 = 0
0r ln 𝑥 = 3 − 2𝑥 in the interval (1,2).
Remark:
If we sketch the two equations 𝑦 = 3 − 2𝑥 𝑎𝑛𝑑 𝑦 = ln 𝑥 we
can estimate the interval on which the solution of the
equation ln 𝑥 = 3 − 2𝑥 (the point of the intersection of the
two curves) lies as shown in the figure below.

𝒚 = 𝟑 − 𝟐𝒙

𝐲 = 𝐥𝐧 𝒙
Example 9
Show that there is a solution to the equation
cos 𝑥 = 𝑥 .
Solution
Let 𝑓 𝑥 = cos 𝑥 − 𝑥 .
Since 𝑓(𝑥) is continuous on the interval [0, 𝜋2] and we have

𝑓 0 =1−0=1 >0
𝜋
and 𝑓 2
= 0 −𝜋2 = −𝜋2 < 0.
𝜋
Since 𝑓 0 > 0 > 𝑓 2

by IVT, there is a number 𝑐 in (0, 𝜋2) such that 𝑓 𝑐 = 0.

Thus, there is a root of the equation cos 𝑥 − 𝑥 = 0,
0r cos 𝑥 = 𝑥 in the interval (0, 𝜋2).
Remark
If we sketch the two equations 𝑦 = 𝑥 𝑎𝑛𝑑 𝑦 = cos 𝑥 we can
estimate the interval on which the solution of the equation
cos 𝑥 = 𝑥 (the point of the intersection of the two curves) lies
as shown in the figure below.
Thank you for listening.