Chapter 1
Limits and Their Properties
1-2 Finding Limits Graphically and
Numerically
• 1-3 Evaluating Limits Analytically
• 1-4 Continuity and One-Sided Limits
• 1-5 infinite Limits
•
This
will test
the
“Limits”
of your
brain!
Sec. 1.4: Continuity
Continuity at a Point
 Properties of Continuity
Continuity on an Open/Closed Interval
 The Intermediate Value Theorem

Importance
Calculus relies heavily upon the existence
of what’s called continuous functions. In
fact, most of the important theorems
require that any function in question
must be continuous before one can
even think about applying the theorems.
“As long as g(x) is a continuous function…”
“Assume h(x) is continuous on [a,b]…”
“Given a continuous function f(x)…”
Intuitive Look at Continuity

A function without
breaks or jumps

The graph can be
drawn without lifting the pencil
Which are continuous throughout?


You can usually tell
by looking if a
function is
continuous at a
certain point.
However, in calculus,
we have a more
formal definition of
continuity.
y = 1/x
y = |x|/x
Continuity at a Point
A function f is continuous at the point x = a if
the following are true:
i ) f (a ) is defined
ii) lim f ( x) exists
xa
iii) lim f ( x)  f (a)
xa
f(a)
a
Discontinuous Graphs
1. f(c) is not defined
hole in graph
x=c
x=c
x=c
2. lim f  x  DNE
x c
jump or asymptote
3. f  c   lim f  x 
x c
x=c
hole in graph and
function defined
elsewhere.
Particular Continuous Functions
 polynomials (monomials)
 f(x) = 2, f(x) = x2 + 1, f(x) = x103 – 4x94 + 7x5
 exponential functions
g(x) = 2ex – 6, g(x) = 5x
 absolute value
h(x) = |x – 3|
sin x and cos x
 ln x, x > 0
 odd root functions
 j(x) =
9
x
even root functions as long as what’s under the radical is positive
 k(x) =
4
x  2 , x > -2
We will now look at 3 functions all of which are discontinuous at some
point and examine which of the 3 properties is violated.
x  2, x  1
1. f(x)  
 1, x  1
i) f 1  1
ii) lim f  x   3
x 1
iii) lim f  x   f (1)
x 1
f(x) is discontinuous at x = 1 because it violates the 3rd property
x2  4
2. f(x) 
x 2
x2  4 (x  2)(x  2)

 x  2,  2
x 2
x 2
Because the factors cancelled out, we know f(x) is a straight line with
a hole in it at x = 2.
Since f(2) is not defined, violating property 1, the discontinuity occurs
at x = 2.
1
 , ifx  0
3. g(x)   x
 1, ifx  0
i) g 1  1
ii) lim g  x  DNE
x 0
g(x) is discontinuous at x = 0 since the limit DNE, which violates
property 2.
You Try…
At which value(s) of x is the given function
discontinuous? (Use your three criteria.)
2
x 9
1. f ( x)  x  2
2. g ( x) 
x3
6
4
2
-4
-2
2
4
-2
f (a)  a  2
═
lim f ( x)  a  2
x a
Continuous everywhere
g (3) is undefined
Continuous everywhere
except at x  3
 x  2, if x  1
3. h( x)  
1, if x  1
 x 2  3x
, if x  0

4. F ( x)   x
1, if x  0

5
4
3
2
1
-2
2
-1
-2
4
h(1)  1
-3
lim h( x)  1 and lim h( x)  3
x 1
x 1

lim h( x) DNE
x 1
Thus h is not cont. at x=1.
F (0)  1

lim F ( x)  3
x 0
Thus F is not cont.
at x = 0.
Two types of discontinuities
1. Removable Discontinuities
Holes: typically caused by a
common factor in the numerator
and denominator – function not
defined violating condition 1
Examples
x 9
g ( x) 
x3
2
Redefined holes: usually
result from a piecewise function
where the equation follows one
rule for every value except one
– violates condition 3
 x2  9
, x  3

g  x   x  3
 -6, x  3

Two Types of Discontinuities
2. Non-removable Discontinuities – violate condition 2
jump
infinite
oscillating
Examples
 x  2, if x  1
h( x )  
1, if x  1
1
g ( x) 
x3
1
j ( x)  sin
x
Non-removable Discontinuities


Vertical Asymptotes - when the denominator is
zero and doesn’t cross out
x2
x2
Ex: 2

x3
x 2  4 ( x  2)( x  2)
hole occurs at x = 2,
but VA at x = -2
Non-removable Discontinuities

Piecewise functions – branches don’t meet
 x  2, if x  1
h( x )  
1, if x  1
5
4
3
2
1
-2
2
-1
-2
-3
4
You Try…
“Discuss the continuity of each function.”
1.
Give any x values where the function is
discontinuous;
2. state the condition for continuity which is violated at
that point;
3. describe what is happening on the graph at each
point; and
4. state whether each continuity is removable or nonremovable.”
x
1) f ( x ) 
2
x2 1
2) f ( x) 
x 1
x 1
3) f ( x)  2
x 1
 x3  1, x  2
4) f ( x)  
 3x, x  2
Proving Continuity


To prove a function is continuous, you must
show ALL 3 conditions are met.
Ex. Prove j(x) is continuous at x = 0.
| x  3 | , if x  0
j ( x)  
2 x  3, if x  0
i) f  0  2(0)  3  3
ii) lim f  x   lim | x  3 | 3
x 0
x 0
x 0
x 0
lim f  x   lim  2 x  3  3
lim f  x   3
x 0
iii) lim f  x   f (0)  3
x 0
You Try…
1. Prove that p(x) is continuous at t = 1.
2t  3t 2
p( x) 
1 t3
2. Is g(x) continuous at x = 2?
g(x)=
5  x , 1  x  2
x 2  1,
2 x3
lim g ( x)  lim 5  x  

x2
x2
3
lim g ( x)  lim x 2  1  3
x2
x2

g(x) is continuous
at x = 2
You Try…
3. What value of a will make f(x) continuous?
 x3 , x  2
f ( x)   2
ax , x  2
Summary:
1.
A discontinuity is REMOVABLE (hole) if the limit exists and it
fails one of the other 2 conditions
*** The hole may or may not be redefined***
2.
A discontinuity is NON-REMOVABLE if it is a jump, vertical
asymptote, or out of the domain of the function.
3.
For piecewise functions, determine whether or not the
branches meet.
4.
A function is continuous at x = c if
1.
f(c) must be defined.
2.
lim f ( x) must exist.
3.
f (c)  lim f ( x )
xc
xc
Properties of Continuity
Properties of continuous functions  If f(x) and g(x) are continuous at x = c, then
the following are continuous at x = c.
1. f(x) + g(x)
2. f(x) . g(x)
.
3. f(x) / g(x), if g(c)  0 4. b f(x)
Also:
 Polynomials are continuous
everywhere.
 Rational functions are continuous
everywhere the
denominator does not
equal 0.
 Radicals are continuous everywhere the
radicand > 0.
Properties of Continuity cont.
Many functions are composite functions so it is important to know that a
composition of continuous functions constructs a continuous function.
Composition: (h ◦ g)(c) or h(g(c))
As a result, the following functions are continuous at every point in their
domains:
2
x
 1 f(x) = 3 tan x
f(x) = x + sin x f (x ) 
cos x
f(x) = sin 4x f ( x )  x  4 f(x) = tan (1/x)
We may use this result to conclude that the inverse trigonometric functions
sin-1(x), cos-1(x), and tan-1(x) are all continuous on their domains.
Continuity on an Open Interval

A function is continuous on an open
interval (a, b) only if it is continuous at
every point in (a, b).
→ If a function is continuous everywhere,
that means it is continuous on the
interval (-∞, ∞).
Determine whether the following functions are
continuous on the given interval.
1.
1
f ( x)  , 0,1
x
yes, it is
continuous
(
( ))
-1
1
What about on the interval (-1, 1)?
NO, it is not continuous since x = 0 makes the
denominator 0.
x 1
f ( x) 
, (0,2)
x 1
2
2.
(
discontinuous at x = 1, so not
continuous on interval
)
3.
f ( x)  sin x, (0,2 )
2
yes, it is continuous on ANY interval
On what interval(s) are the following functions
continuous?
1
1. h( x ) 
x
x 1
2. g ( x )  2
x 1
3.
(-∞, 0), (0, ∞)
(-∞, -1), (-1,1), (1, ∞)
 x  1, x  0
f ( x)   2
 x  1, x  0
(-∞, ∞)
Continuity on a Closed Interval [a,b]
A function is continuous on a closed interval
[a,b] if it continuous on the open interval
(a, b) and
lim f ( x)  f (a)
x a
lim f ( x)  f (b)
x b
a
b

Ex: p( x)  x  1  2
is continuous on [1, ∞) since p(1) = 2
and lim p( x)  2 .
x 1
Example


Is f ( x)  9  ( x  5) 2 continuous on the
interval [2, 8]?
Yes, since lim f ( x)  0 and f (2)  0 ,
x2
plus lim f ( x)  0 and f (8)  0 .

x 8
Continuity on an Interval

On what intervals are the following functions
continuous?
2. g ( x)  sin x
x2 1
1. f ( x)  2
x 4
4. f ( x)  x x  3
x2
2 x,
3. f ( x)   2
 x  4 x  1, x  2
Intermediate Value Theorem (intuitively)
Suppose that a boy is 5 feet tall on his thirteenth birthday and
5 feet 7 inches tall on his fourteenth birthday. Then, for any
height h between 5 feet and 5 feet 7 inches, there must have
been a time t when his height was exactly h.
This seems reasonable because human growth is continuous
and a person’s height does not abruptly change from one
value to another.
Intermediate Value Theorem
If f is a continuous function on a closed interval [a, b] and
L is any number between f (a) and f (b), then there is at
least one number c in [a, b] such that f(c) = L.
y  f ( x)
f (b)
What would it mean
if f(a) and f(b) have
opposite signs?
f (c) = L
f (a)
a c
b
The Intermediate Value Theorem
There may be multiple x-values
such that f(c) = k.
A function that is not continuous
does not necessarily exhibit the
intermediate value property.
y
8
7
6
Pick a closed interval on the x axis say [1,2.5]
5
4
f(1) = 3 and f(2.5) = 5
3
So if I pick a y value between 3 and 5 (say 4),
then there must be a value c between 1 and 2.5
such that f(c) = 4 c = 2.25.
2
1
9 8 7 6 5 4 3 2 1
1
In this case, c is approximately 2.3.
2
3
4
5
6
7
1
2
3
4
5
6
Use Intermediate Value Theorem to verify that the
function f(x) = x3 – x2 + x – 2 equals 4 for some c
on the interval [0, 3].
Solution: f(x) is continuous on the interval [0, 3]
(and everywhere else for that matter), and
f(0) = -2
f(3) = 19
Therefore by the IVT somewhere between x = 0, and
x = 3, there exists a c such that f(c) = 4.
Verify the IVT applies and find all values guaranteed
by the theorem.
h (x )  x  2x ; [1, 4]  k  8
2
MUST:
1. State f(x) is continuous on [a, b].
2. Find f(a) and f(b)
3. Find x = c such that f(c) = k
You Try…
(a) Verify that the Intermediate Value
Theorem applies in the indicated interval
for f(x). (b) Find the value of c
guaranteed by the theorem.
2x
1) g (x ) 
; [0, 6]  k  1
x 1
4
2) f (x ) 
; [0, 4]  k  2
x 3
Locating Roots with IVT

Given f(a) and f(b) have opposite signs,
then there must be a root between a
and b.
f(a)
a
b
f(b)
Given f ( x)  3 x  2 x  5, show that
2
f ( x)  0 has a solution on 1, 2.
f (1)  4  0
f (2)  3  0
f (x) is continuous and since f (1) < 0 and
f (2) > 0, by the Intermediate Value Theorem
there exists a c on [1, 2] such that f (c) = 0.
For the function f(x) = x2 -6x + 8 , find the value of "c"
on [0, 3] guaranteed by the IVT such that f(c) = 0.
f(0) = 8
f(3) = -1
0 is between -1 and 8.
f(c) = 0
c2 – 6c + 8 = 0
(c – 4)(c – 2) = 0
c = 2 or 4
Not 4, why?
You Try…
1. Use the Intermediate Value Theorem to show
that the polynomial function f(x) = x3 +2x – 1 has a
zero in the interval [0, 1].
2. Given g(x) = x3 – x – 1 [1, 2], find the value of c
guaranteed by the IVT such that g(c) = 0.
Solution:
f is continuous on the closed interval [0, 1].
Because f(0) = -1 and f(1) = 2, it follows that
f(0) < 0 and f(1) > 0.
You can therefore apply the
Intermediate Value
Theorem to conclude that
there must be some c in
[0, 1] such that f(c) = 0.