# 2.5 Continuity

advertisement ```MATH 151: 2.5 Continuity
Informally speaking, if a function is continuous, then you do not have to lift your pencil
when drawing the graph from left to right. If you have to lift your pencil off the page
when x = c , then we say that the function is discontinuous at x = c . Identify which
functions below are discontinuous, where and in what way.
Formally speaking, there are three conditions that must be true in order for a function
to be continuous at x = c .
Definition of Continuity: A function f is continuous at x = c if
()
1. the function value f c is defined or exists. This requires that
a) c ∈Df
()
b) f c is a single, real and finite value
()
2. the limit value lim f x is defined or exists. This requires that
x →c
()
()
a) the left and right limits are equal: lim f x = lim f x (therefore no jumps)
()
x →c −
x →c +
b) the limit value is finite: lim f x ≠ ±∞ (therefore no vertical asymptotes)
x →c
() ()
3. the limit value equals the function value: lim f x = f c . This means that
x →c
a) the height that y approaches as x approaches c from any direction is the same
height that y equals when x equals c.
b) there are no jumps, holes, vertical asymptotes or oscillating discontinuities
If x = a is the left endpoint of an interval on which f is defined, then f is continuous at
() ()
x = a if lim f x = f a .
x →a +
If x = b is the right endpoint of an interval on which f is defined, then f is continuous
() ()
at x = b if lim f x = f b .
x →b +
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MATH 151: 2.5 Continuity
Consequence of Continuity: If a function is continuous at x = c , then (direct)
()
substitution can be used to evaluate lim f x .
x →c
Exercise 1: Define
a) Domain
b) Interval of Continuity
For some functions, the domain and intervals of continuity are the same, such as
polynomials ( x ∈ ) and rational functions ( x ∈ except for the x – values that make
the denominator zero, making the function value indeterminate OR undefined and
thereby producing a hole OR a vertical asymptote).
Exercise 2: Give an example of one function that has a different domain from its
interval of continuity.
⎧2x + c , x < 4
⎪
Exercise 3: Solve for c such that f x = ⎨
is continuous at x = 4 .
⎪⎩4 x − 1, x ≥ 4
()
⎧
π
⎪x sin x , x <
⎪
4 is continuous at x = π .
Exercise 4: Solve for a such that f x = ⎨
4
⎪ax 2 − 3a, x ≥ π
⎪⎩
4
()
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MATH 151: 2.5 Continuity
()
Exercise 5: Use the graph of y = f x below to complete the table of function and limit
()
values below. Then, conclude whether or not y = f x is continuous at each of the x –
values specified, and state the domain and interval of continuity.
x = c:
x = −3
x =−
6
5
x =0
x =2
x =4
()
f c =?
()
lim f x = ?
x →c
() ()
Is f (x )
lim f x = f c ?
x →c
continuous
@ x = c?
State the domain of the function:
State the interval of continuity of the function:
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MATH 151: 2.5 Continuity
Types of Discontinuities
If at least one of the three Continuity Conditions fails, then the function is
discontinuous at x = c and has either a REMOVABLE discontinuity (hole) or a NONREMOVABLE discontinuity (infinite, jump or oscillating) at x = c .
The function has a non-removable discontinuity if:
CASE I: Infinite Discontinuity
CASE II: Jump Discontinuity
The function value is UNDEFINED The right and left limits are unequal,
(vertical asymptote).
lim f x ≠ lim f x ; thus lim f x = ∃ .
()
x →c +
non-removable Example:
y=
x →c −
1
has
x
discontinuity at x = 0 .
Example:
y=
a
()
x →c
x
has
x
discontinuity at x = 0 .
a
()
non-removable
The function has a removable discontinuity if:
CASE I: The function value is NOT defined CASE II: The function and limit values are
at all (as in the case of a hole).
different: f c ≠ lim f x .
()
Example:
(x − 1)
y=
2
x −1
discontinuity at x = 1 .
has
© Raelene Dufresne 2013
a
x →c
()
⎧x − 1, x ≠ 1
y
=
Example:
has a removable
⎨3,
removable
x =1
⎩
discontinuity at x = 0 .
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MATH 151: 2.5 Continuity
Oscillating Discontinuity:
If the function oscillates too much to have a limit as
x → c , the type of (non-removable) discontinuity is
called an oscillating discontinuity
Example: y = sin
1
x
has an oscillating discontinuity at
⎛ 1⎞
x = 0 , therefore lim sin ⎜ ⎟ = ∃ .
x →0
⎝x⎠
Removing a Discontinuity:
Removable discontinuities are discontinuities that can be “removed”: the function can be
extended by redefining the function as a piecewise function with a new function value at
()
()
the discontinuity x = c , such that f c = lim f x .
Example:
()
lim f x
x →2
()
f2
Determine
and
from the graph at the
right. Then determine if the
discontinuity at x = 2 is removable
or non-removable.
If it is
removable, define or redefine the
function value to make the function
continuous.
x →c
⎧1
⎪ x, x ≠ 2
f x = ⎨2
⎪−1, x = 2
⎩
()
Solution:
()
Continuity Condition 2 holds: The limit value exists, lim f (x ) = 1 .
Continuity Condition 1 holds: The function value exists, f 2 = −1 .
x →2
()
()
Continuity Condition 3 fails: The function value equals the limit value: f 2 ≠ lim f x .
()
x →2
So, in order to “remove the discontinuity”, the function value f 2 must be redefined to
()
()
equal the limit value lim f x = 1 (so that the “point” fills the “hole”): Define f 2 = 1 .
x →2
⎧1
1
⎪ x, x ≠ 2
Therefore f x = ⎨ 2
, or simply, f x = x , ∀x ∈ .
2
⎪1, x = 2
⎩
()
© Raelene Dufresne 2013
()
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MATH 151: 2.5 Continuity
Exercise 6: Identify the location of the discontinuity and remove it for the function
()
f x =
x 2 − 4x − 5
x2 − 1
. Sketch the original and the extended function.
You might ask why we study the continuity of functions. Though the point is subtle,
because we are used to studying functions that are mostly continuous on the entire
domain, the point is important:
Continuous functions map connected sets of x – values to connected sets of y – values.
()
y =f x
()
f (c )
f (a )
f b
()
The function y = f x
is continuous at
x = c , which means that a connected set
of x – values containing x = c , namely
x ∈ ⎡a, b ⎤
⎣
⎦
will get mapped by the continuous
function f to a connected set of y –values
()
containing f c , namely
() ()
y ∈ ⎡⎢f a , f b ⎤⎥ .
⎣
⎦
a c b
Intermediate Value Theorem for Continuous Functions
A function that is continuous on a closed interval x ∈ ⎡a, b ⎤ takes on every value
⎣
⎦
( ) and f (b ) .
between f a
( ) and f (b ) for some x
In other words, if y o is between f a
()
value c ∈ ⎡a, b ⎤ , then y o = f c .
⎣
⎦
Make a sketch to illustrate the Intermediate Value Theorem.
Exercise 7: Given values for a continuous function f, where MUST f have x – intercepts?
Explain.
x
()
f x
-5
-3.2
-2.9
-0.8
1.2
1.23
7
0
-100
-2
3
1
-0.1
-2
Exercise 8: Is any real number exactly 1 less than its cube?
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