Calculus I: Continuity of function
1. Definition of continuity
A function f(x) is said to be continuous at a point (a, f(a))
if
each of the following conditions is satisfied:
(1) f(a) is defined
(2) lim f(x) exists
x → a
(3) lim f(x) = f(a)
x → a
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If "a" is a point in the domain of the function f, then
the function f is continuous at x = a if and only if
lim f(x) = f(a)
x → a
A function f(x) is continuous if it is continuous at every point of its domain.
The graph of f is a connected curve with no jumps or gaps.
2. Examples:
2.1. Example 1:
The function
f(x) = x2 + 1
is continuous in all its domain; that is
]- ∞, + ∞[
2.2. Example 2:
The function
f(x) = 1/(x - 2) is not continuous at the point x = 2;
because f(2) is not defined.
Therefore is discontinuous at the point x = 2.
2.3. Example 3:
The function
f(x) = (x2 - 4)/(x - 2) is not continuous at the point x = 2;
because the function is not defined at the point x = 2.
(x2 - 4)/(x - 2) is not continuous at the point x = 2.
2.4. Example 4:
The function
f(x)
= + 2 if x >= 0
= - 2 if x < 0
We have f(2) = 0 , and
lim f(x) = 2
x → 0+
and
lim f(x) = - 2
x → 0-
so
lim f(x)
x → 0
The second condition is not satisfied, then
f is discontinuous at the point x = 0.
2.5. Example 5:
Determine if the following function is continuous at x = 1:
f(x) =
2 x - 1, if x ≠ 1
3 , if x = 1
(1) f(1) = 2(1) - 1 = 1, so
f(1) exists , then the function f is defined at x = 1.
(2) lim f(x) = 3, exists
x → 1
(3) lim f(x) ≠ f(a)
x → a
The condition (3) is not satisfied, so the
function f is not continuous at x = 1
3. Left and right continuity
3.1. Definitions
We have the following properties:
A function f is left continuous of x = a
if and only if
lim f(x) = f(a)
x → a-
A function f is right continuous of x = a
if and only if
lim f(x) = f(a)
x → a+
3.2. Example
f(x) = x - 3 if x > 3
f(x) = - (x - 3) if x < 3
f(x) = 0 if x = 3
We now consider the limits of f as x approaches 3 from
the left (x < 3) where f(x) = - (x - 3):
lim f(x) = lim - (x - 3) = 0
x → 3-
We now consider the limits of f as x
approaches 3 from the right (x > 3) where f(x) = + (x - 3).
lim f(x) = lim + (x - 3) = 0
x → 3+
The two limits are equal. So
lim f(x) = 0
x → 3
Then this limit exists and is equal to 0 .
Since
lim f(x) = 0 = f(3)
x → 3
the function f is continuous at x = 3.
(and also for all values of x in R).
4. Continuity on open and closed interval
4.1. Definitions
We have the following properties:
A function f is continuous on an open interval ]a, b[
if the function f is continuous at any point
within this open interval.
A function f is continuous on an closed interval [a, b]
if the function f is :
a) continuous on the open interval ]a,b[
b) right continuous at a
c) left continuous at b
4.2. Example
The function
f(x) = (x - 2)1/2 is
- not left continuous at x = 2
- right continuous at x = 2
so
The function f is continuous on [2, + ∞[
5. Other properties
5.1. Properties
We have the following properties:
Any polynomial function is continuous on the whole real set R.
Any rational function is continuous on its domain.
Any irrational function of the form f(x) = [g(x)]1/n
(n  N) is continuous on its domain.
We may exclude the points of discontinuity
on the bounds of this domain.
I f and g are two functions continuous on the interval
I, then:
a) kf is continuous on I (k R)
b) f ± g is continuous on I,
c) f . g is continuous on I ,
d) f/g is continuous on I (except the values of
the interval I which cancel the denominator)
5.2. Examples
The function
f(x) = (x - 2)1/2 is
- not left continuous at x = 2
- right continuous at x = 2
so
The function f is continuous on ]2, + ∞[
3. Exercises:
Determine the continuity of the following
function:
a)
f(x) = (x2 - 16)/(x - 4)
at the point x = 0,
at the point x = 4
at the point x = 5
b)
f(x) = √(x - 3)
at the point x = 0,
at the point x = 3
at the point x = 3+
at the point x = 3-
c)
f(x) = √(x - 3)
at the point x = 0,
at the point x = 3
at the point x = 3+
at the point x = 3-
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