HIGH SCHOOL
DIFFERENTIAL CALCULUS
COURSE
3.4. CONTINUITY OF A FUNCTION
3.1.1 Define the continuity of a function at a point
3.1.2 Determine if a function is continuous or discontinuous at a given point from
its algebraic expression or its graph
Let f be a function and c  Domain( f ) . We say that f is continuous at x  c if and only if
lim f ( x)  f (c) . Knowing that if lim f ( x)  lim f ( x)  L it can be said that lim f ( x)  L , so the
x c
x c
x c
x c
condition of a function to be continuous is lim f ( x)  lim f ( x)  lim f ( x)  f (c)  L .
x c
x c
x c
If the function f is not continuous at x=c, then f is said to have a discontinuity at c. There are two
types of discontinuties.
- Removable discontinuity
- Nonremovable discontinuity
o Jump discontinuity
o End of domain
o Infinite discontinuity
1) Removable discontinuity is the discontinuity at a point which could be removed by defining the function
at just that one point. This often appears as a graph with a hole in it(gap). A removable discontinuity it
appears when
An example of a removable discontinuity could be considered the function f ( x) 
x2  4
in the point x=2.
x2
 x  2 x  2  lim x  2  4
x2  4  0 
lim
    lim


x 2 x  2
x  2
x2
 0  x 2
lim
 x  2 x  2  lim x  2  4
x2  4  0 
    lim


x  2
x  2  0  x 2
x2
lim
x2  4
4
x2
x 2
x 2
f(2)=dne
x2  4
x2  4
x2  4
 lim
 lim
 4 and
so because lim
x 2 x  2
x 2 x  2
x 2 x  2
f(2)=dne, the function is not continuous in x=2 and it
has a removable discontinuity.
 x2  4
if x  2

f ( x)   x  2
in x=2
2
if x  2

 x  2 x  2  lim x  2  4
x2  4  0 
lim
    lim


x 2 x  2
x  2
x2
 0  x 2
lim
x 2
 x  2 x  2  lim x  2  4
x2  4  0 
    lim


x  2
x  2  0  x 2
x2
so lim
x 2
x2  4
4
x2
f(2)=2
so because
x2  4
x2  4
x2  4
 lim
 lim
 4 and
x 2 x  2
x 2 x  2
x 2 x  2
f(2)=2, the function is not continuous in x=2 and it has
a removable discontinuity.
2) Nonremovable discontinuity is the discontinuity at a point which could not be removed at just that point.
There are three types of nonremovable discontinuities:
2a) Jump discontinuity is usually caused by a piecewise-defined function whose pieces don't really
meet together.
lim
In other words lim f ( x)  L1 and lim f ( x)  L2 and L1  L2
x c
x c

 x  2 if x   ,1
Example1: f ( x)  
 x  2 if x  1, 4
lim   x  2   1  2  3
x 1
lim
x 1

 
x 2 

1  2  1
.
because the limit from the left is not equal with the
limit from the right
lim f ( x)  dne
x 1
and f(1)= -1
The function is not continuous in the x=1 and it has
a jump discontinuity
2b) End of domain is when the function has a limit from one side, and from the other side does not
exist.
y
In other words
lim f ( x)  L
x c
and

lim f ( x)  dne

x c

OR

x
lim f ( x)  dne and
x c
lim f ( x)  L
x c













An example of end of domain discontinuity is: f ( x) 
x  2 1
lim

x  2  1  0  1  dne
lim

x  2  1  0  1  1
x2
x2
so lim
x 2




x  2  1  dne
The value of the function in x=2 is f(2)=1
Because the limit does not exist from the left
and the limit from the right is 1, the type of
discontinuity is end of domain.
2c) Infinite discontinuity is defined where the jump is to the infinity
In other words
lim f ( x)   and lim f ( x)  
x c 
x c
OR
lim f ( x)   and lim f ( x)  
x c 
x c
y




x









f ( x) 
1
in x=0
x




y

1  1
lim     
x 0  x 
0

  

1  1
lim     
x 0  x 
0

  




x









1 1
lim       dne
x 0 x
  0

1
f (0)     dne
0


Because the limit from the left is not equal with
the limit from the right, the function is not
continuous in x=0 and the type of discontinuity
is infinite discontinuity.

Discontinuity at a point.
A function has a discontinuity at a point x=c if and only if lim f ( x)  f (c)
x c
Discontinuity on an interval
A function f is continuous on the closed interval  a, b if it is continuous on the open
interval  a, b  and lim f ( x)  f (a) and lim f ( x)  f (b)
x a
x b
Example1:
Given the following piecewise function
2 x  1
if x   , 2

f ( x)  1  x 2
if x   2, 2 

 x  2  1 if x   2,  
a) Analyze the function in each step point and decide the continuity or the type of
discontinuity( if it exists)
b) Decide if the function is continuous in its entirety
c) Graph the function
d) Range of the function
a) To analyze the function in its step-points, it is needed to check the continuity in
these step points (Step-points are the points of a piecewise function, where a
component function is finishing, and other component function is starting). In our
case, there are two step-points where is needed to check the continuity, at x=-2
and x=2.
In x=-2
From the left of x=-2, lim (2 x  1)  2   2   1  4  1  3 and from the right,
x 2

lim (1  x )  1  (2)
2
x 2
2
  1  4  3 . Because the limit from the left and the limit
from the right has the same value, it can be decided that lim f ( x)  3
x 2
The value of the function at x=-2 is f (2)  2   2  1  4  1  3 .
Because of lim (2 x  1)  f (2)  3 it can be concluded that the piecewise function
x 2
is continuous in x=-2
In x=2
From the left of x=2, lim (1  x 2 )  1  22  3 and from the right,
lim
x  2

 
x  2 1 
x 2



2  2  1  2  1  3 . Because the limit from the left and the limit
from the right has different values, in this point there is a discontinuity. So, the
limit lim f ( x)  dne .
x2
The value of the function at x=2 is f (2)  2  2  1  4  1  2  1  3 .
The limit lim f ( x)  dne because of the limit from the left has a different value
x2
than the limit from the right and it can be concluded that the piecewise function
has a jump discontinuity at x=-2
b) To decide the continuity in its entirety, it is needed to check each function´s
continuity. The first function, f ( x)  2 x  1 for x   , 2 is continuous in its entirety
of domain, the second function f ( x)  1  x2 for x   2, 2 also is continuous in its
entirety of domain and the third function f ( x)  x  2  1 for x   2,   is also
continuous in its domain.
d)
y


The range of the function is:

y   ,1

x













3, 