Sec 2.5: CONTINUITY. Study continuity at x = 4 Sec 2.5: CONTINUITY Study continuity at x = 2.
Section 1.4 – Continuity and One-Sided Limits
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Transcript of Section 1.4 – Continuity and One-Sided Limits
Section 1.4 – Continuity and One-Sided Limits
ExampleFind values of a and b that makes f(x) continuous.
f x ax 3 if x 58 if x 5
x 2 +bx +1 if x 5
When x=5, all three pieces must have a limit of 8.
5a 5
ax 3 8
a 5 3 8
a 1
5b 28 8
x 2 bx 3 8
5 2 b 5 3 8
5b 20
b 4
Continuity at a Point
A function f is continuous at c if the following three conditions are met:
1. is defined.
2. exists.
3.
( )f c
lim ( )x c
f x
lim ( ) ( )x c
f x f c
c
L
f(x)
x
For every question of this type, you need (1), (2), (3), conclusion.
Example 1
Show is continuous at x = 0.
f x 2 1 x 2
1. f 0
2 1 02
1The function is clearly
defined at x= 0
2. limx 0
2 1 x 2
2 1 02
1With direct
substitution the limit clearly exists at x=0
3. f 0 limx 0
2 1 x 2The value of the
function clearly equals the limit at x=0
f is continuous at x = 0
Example 2
Show is not continuous at x = 2. 8 1 if 210 if 2
x xf x
x
1. 2f 10 The function is clearly 10 at x = 2
2
lim 8 1x
x
8 2 1 15
With direct substitution the limit clearly exists at x=0
2
3. 2 lim x
f f x
The value of the function clearly does not equal the limit at x=2
f is not continuous at x = 2
The behavior as x approaches 2 is dictated by 8x-1
2
2. lim x
f x
DiscontinuityIf f is not continuous at a, we say f is discontinuous
at a, or f has a discontinuity at a.
Typically a hole in the curve
Step/Gap Asymptote
Types Of DiscontinuitiesRemovable
Able to remove the “hole” by defining f at one point
Non-RemovableNOT able to remove the “hole” by defining f at
one point
ExampleFind the x-value(s) at which is not
continuous. Which of the discontinuities are removable?
22 153( ) x x
xf x
There is a discontinuity at x=-3 because this makes the
denominator zero.
If f can be reduced, then the discontinuity is removable:
2 5 33
x xxf x
2 5 33
x xx
2 5x This is the
same function as f except at
x=-3f has a removable
discontinuity at x = -3
Notice that:
003f
Indeterminate Form: 0/0
Let:
g xf x
h x
If:
00
g cf c
h c
Then f(x) has a removable discontinuity at x=c.
If f(x) has a removable discontinuity at x=c.
Then the limit of f(x) at x=c exists.
c
L
f(x)
x
One-Sided Limits: Left-HandIf f(x) becomes arbitrarily close to a single REAL
number L as x approaches c from values less than c, the left-hand limit is L.
The limit of f(x)…
as x approaches c from the left…
is L.Notation:
c
L
f(x)
x
lim ( )x c
f x L
lim ( )x c
f x L
One-Sided Limits: Right-HandIf f(x) becomes arbitrarily close to a single REAL
number L as x approaches c from values greater than c, the right-hand limit is L.
The limit of f(x)…
as x approaches c from the right…
is L.Notation:
c
L
f(x)
x
Example 1Evaluate the following limits for
f x x
limx 1
f x
limx 1
f x
limx 1
f x
1
0
DNE
3.5
limx
f x
3.5
limx
f x
3.5
limx
f x
3
3
3
Example 2
Analytically find . 12
24
3 if 4lim if
if 4x
x xf x f x
x x
12If is approaching 4 from the left, the function is defined by 3x x
4
limx
f x
124
lim 3x
x 1
2 4 3
4
Therefore lim 5x
f x
5
The Existence of a Limit
Let f be a function and let c be real numbers. The limit of f(x) as x approaches c is L if and only if
lim ( ) lim ( )x c x c
f x L f x
c
L
f(x)
x
Right-Hand LimitLeft-Hand Limit =A limit exists if…
Example 1Analytically show that .
limx 2
x 2 1 1
Evaluate the right hand limit at 2 :x
limx 2
x 2 1
limx 2
x 2 1
2 2 1
2 1
x
if 2if 2
xx
1
Evaluate the left hand limit at 2 :x
limx 2
x 2 1
limx 2
x 2 1
2 2 1
1
Therefore limx 2
x 2 1 1
Use when x>2
Use when x<2You must use the piecewise equation:
2 +12 +1
xx
Example2Analytically show that is continuous at x = -
1. = 1f x x
Evaluate the right hand limit at 1:x
1lim 1
xx
1lim 1
xx
1 1
f x
if 1if 1
xx
0
Evaluate the left hand limit at 1:x
1lim 1
xx
1lim 1
xx
1 1 0
Therefore is continuous at 1f x x
Use when x>-1
Use when x<-1
You must use the piecewise equation:
11
xx
1. 1f 1 1 0
1
2. Find limx
f x
0
1
3. 1 lim x
f f x
Continuity on a Closed Interval
A function f is continuous on [a, b] if it is continuous on (a, b) and
lim ( ) ( ) lim ( ) ( )x a x b
f x f a and f x f b
a
f(a)
f(b)
xb
Must have closed dots on the endpoints.
t x
Example 1Use the graph of t(x) to determine the intervals on
which the function is continuous.
6, 3 3,0 0,2 2,5 5,6
Example 2Discuss the continuity of
f x 1 1 x 2
The domain of f is [-1,1]. From our limit properties, we can say it is continuous on (-
1,1)
By direct substitution:
limx 1
f x
1 1 1 2
1
f 1
limx 1
f x
1 1 1 2
1
f 1
Is the middle is continuous?
Are the one-sided limits of the endpoints equal to the functional value?
f is continuous on [-1,1]
Properties of ContinuityIf b is a real number and f and g are continuous at x = c,
then following functions are also continuous at c:1. Scalar Multiple:
2. Sum/Difference:
3. Product:
4. Quotient: if
5. Composition:
Example: Since are continuous, is continuous too.
f og x
f x g x
f x g x
b f x
f x g x
g c 0
f x 2x and g x x 2
h x 2x x 2
Intermediate Value TheoremIf f is continuous on the closed interval [a, b] and k
is any number between f(a) and f(b), then there is at least one number c in [a, b] such that:
( )f c k
a b
f(a)
f(b)
k
c
This theorem does NOT find
the value of c. It just proves it
exists.
Free Response Exam 2007
1h 1 6f g 2 6f 9 6 3
3h 3 6f g 4 6f 1 6 7
Since h(3) < -5 < h(1) and h is continuous, by the IVT, there
exists a value r, 1 < r < 3, such that h(r) = -5.
Notice how every part of the theorem is
discussed (values of the function AND continuity).
We will learn later that this implies continuity.
ExampleUse the intermediate value theorem to show
has at least one root.
f x 4 x 3 6x 2 3x 2
f 0 4 0 3 6 0 2 3 0 2
2
f 2 4 2 3 6 2 2 3 2 2
12Find an output greater than zero
Find an output less than zero
Since f(0) < 0 and f(2) > 0
There must be some c such that f(c) = 0 by the IVTThe IVT can be used since f
is continuous on [-∞,∞].
ExampleShow that has at least one solution on the
interval .
cos x x 3 x
f 4 cos
4 4 3
4
1.008
f 2 cos
2 2 3
2
2.305Find an output less than zero
Find an output greater than zero
Since and
There must be some c such that cos(c) = c3 - c by the IVT
The IVT can be used since the left and right side are
both continuous on [-∞,∞].
4 ,
2 Solve the equation for zero.
cos x x 3 x 0
f x
f 4 0
f 2 0