2.2 Limits Involving Infinity
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4
1
f  x 
x
3
2
1
-4
1
lim  0
x  x
-3
-2
-1
0
1
2
3
4
-1
-2
-3
-4
As the denominator gets larger, the value of the fraction
gets smaller.
There is a horizontal asymptote if:
lim f  x   b
x 
or
lim f  x   b
x 

Example 1:
lim
x 
x
x2  1
 lim
x 
x
x2
x
 lim
1
x  x
This number becomes insignificant as

x  .
There is a horizontal asymptote at 1.

sin x
f  x 
x
Example 2:
Find:
sin x
lim
x 
x
2
1
-12 -10 -8
-6
-4
-2 -10
-2
1  sin x  1
so for x  0 :
2
6
8
10 12
When we graph this
function, the limit appears
to be zero.
1 sin x 1


x
x
x
1
sin x
1
lim  lim
 lim
x  x
x 
x  x
x
sin x
0  lim
0
x 
x
4

by the sandwich
theorem:
sin x
lim
0
x 
x

Example 3:
Find:
5 x  sin x
lim
x 
x
 5 x sin x 
lim  

x   x
x 
sin x
lim 5  lim
x 
x 
x
50
5

Infinite Limits:
4
3
1
f  x 
x
2
1
-4
As the denominator approaches
zero, the value of the fraction gets
very large.
-3
-2
-1
0
If the denominator is negative then
the fraction is negative.
2
3
4
-1
-2
-3
-4
If the denominator is positive then the
fraction is positive.
1
vertical
asymptote
at x=0.
1
lim  
x 0 x
1
lim  
x 0 x

Example 4:
1
lim 2  
x 0 x
1
lim 2  
x 0 x
The denominator is positive
in both cases, so the limit is
the same.
1
 lim 2  
x 0 x

End Behavior Models:
End behavior models model the behavior of a function as
x approaches infinity or negative infinity.
A function g is:
a right end behavior model for f if and only if
f  x
lim
1
x  g  x 
a left end behavior model for f if and only if
f  x
lim
1
x  g  x 

Example 7:
f  x  x  e
x
x
x


e
As
,
approaches zero. (The x term dominates.)
 g  x   x becomes a right-end behavior model.
Test of
model
xe
lim
x 
x
x
e x
 1 0  1
 lim1 
x 
x
Our model
is correct.
x
As x  , e increases faster than x decreases,
x
therefore e is dominant.
 h  x   e x becomes a left-end behavior model.
x
Test of
Our model
x
x

e
model

lim

1
 0  1  1 is correct.
lim

x

x
x

x 
e
e

Example 7:
f  x  x  e
x
 g  x   x becomes a right-end behavior model.
 h  x   e x becomes a left-end behavior model.
On your calculator, graph:
y1  x
y2  e  x
y3  x  e  x
Use:
10  x  10
1  y  9

Example 7:
2 x5  x 4  x 2  1
f  x 
3x 2  5 x  7
Right-end behavior models give us:
2 x5 2 x3

2
3x
3
dominant terms in numerator and denominator

Often you can just “think through” limits.
1
lim sin  
x 
x
0
 lim sin x
x 0
0
p
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2.3 Continuity
Most of the techniques of calculus require that functions
be continuous. A function is continuous if you can draw it
in one motion without picking up your pencil.
A function is continuous at a point if the limit is the same
as the value of the function.
This function has discontinuities
at x=1 and x=2.
2
1
1
2
3
4
It is continuous at x=0 and x=4,
because the one-sided limits
match the value of the function

Removable Discontinuities:
(You can fill the hole.)
Essential Discontinuities:
jump
infinite
oscillating

Removing a discontinuity:
x3  1
f  x  2
x 1
has a discontinuity at x  1 .
Write an extended function that is continuous
at x  1 .
 x  1  x 2  x  1 1  1  1
x3  1
 lim

lim 2
x 1
x 1 x  1
 x  1 x  1
2
 x3  1
 2 , x  1
f  x    x 1
 3 , x 1
 2

3
2
Note: There is another
discontinuity at x  1 that can
not be removed.
Removing a discontinuity:
5
4
3
2
1
-5 -4 -3 -2 -1 0
-1
1
2
3
4
5
-2
-3
-4
-5
 x3  1
 2 , x  1
f  x    x 1
 3 , x 1
 2
Note: There is another
discontinuity at x  1 that can
not be removed.

Continuous functions can be added, subtracted, multiplied,
divided and multiplied by a constant, and the new function
remains continuous.
Also: Composites of continuous functions are continuous.
examples:
y  sin  x 2 
y  cos x

Intermediate Value Theorem
If a function is continuous between a and b, then it takes
on every value between f  a  and f  b  .
f b
Because the function is
continuous, it must take on
every y value between f  a 
and f  b  .
f a
a
b

Example 5:
Is any real number exactly one less than its cube?
(Note that this doesn’t ask what the number is, only if it exists.)
f 1  1
x  x3  1
0  x3  x  1
f  x   x3  x  1
f  2  5
Since f is a continuous function, by the
intermediate value theorem it must
take on every value between -1 and 5.
Therefore there must be at least one
solution between 1 and 2.
Use your calculator to find an approximate solution.
solve  x  x 3  1, x 
F2
1: solve
1.32472

Graphing calculators can sometimes make noncontinuous functions appear continuous.
Graph:
y  floor  x 
CATALOG
Note resolution.
F
floor(
This example was graphed
on the classic TI-89. You
can not change the
resolution on the Titanium
Edition.
The calculator “connects the dots”
which covers up the discontinuities. 
Graphing calculators can make non-continuous
functions appear continuous.
Graph:
y  floor  x 
CATALOG
F
floor(
If we change the plot style
to “dot” and the resolution
to 1, then we get a graph
that is closer to the
correct floor graph.
The open and closed circles do not
show, but weGRAPH
can see the
discontinuities.
p
2.4 Rates of Change and Tangent Lines
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The slope of a line is given by:
y
m
x
y
x
The slope at (1,1) can be approximated by
the slope of the secant through (4,16).
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
y 16  1 15


5
x
4 1
3
We could get a better approximation if we
move the point closer to (1,1). ie: (3,9)
y
9 1
8


x
3 1
2
4
0 1 2 3 4
f  x  x
Even better would be the point (2,4).
2
y
4 1
3


x
2 1
1
3

The slope of a line is given by:
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
y
m
x
y
x
If we got really close to (1,1), say (1.1,1.21),
the approximation would get better still
y 1.21  1 .21


 2.1
.1
x 1.1  1
How far can we go?
0 1 2 3 4
f  x  x
2

y
slope 
x
f 1  h 
f 1
slope at 1,1
h
1 1 h
f 1  h   f 1

h
1 h

 lim
h 0
2
1
h
h  2  h
1  2h  h 2  1
 lim
 lim
h 0
h 0
h
h

2

The slope of the curve y  f  x  at the point P a, f  a  is:
f  a  h  f  a
m  lim
h 0
h



The slope of the curve y  f  x  at the point P a, f  a  is:
f  a  h  f  a
m  lim
h 0
h
f a  h  f a
h
is called the difference quotient of f at a.
If you are asked to find the slope using the definition or using
the difference quotient, this is the technique you will use.

The slope of a curve at a point is the same as the slope of
the tangent line at that point.
In the previous example, the tangent line could be found
using y  y1  m  x  x1  .
If you want the normal line, use the negative reciprocal of
the slope. (in this case, 
1
)
2
(The normal line is perpendicular.)

Example 4:
1
Let f  x  
x
a Find the slope at x  a .
On the TI-89:
limit ((1/(a + h) – 1/ a) / h, h, 0)
F3
Calc
f  a  h  f  a
m  lim
h 0
h
1 a  a  h 1 a  a  h

a
 lim a  h
h 0
h a  a  h
1 a  a  h
 lim 
h 0 h
a a  h
aah
 lim
0
h 0 h  a  a  h 
Note:
If it says “Find the limit”
on a test, you must
show your work!
1
 2
a

Example 4:
On the TI-89:
1
Let f  x  
x
b Where is the slope  1 ?
4
1
1
  2
4
a
a2  4
a  2
Y=
y=1/x
WINDOW
6  x  6
3  y  3
x scl  1
y scl  1
GRAPH

Example 4:
1
Let f  x  
x
b Where is the slope  1 ?
4
We can let the calculator
Ontangent:
the TI-89:
plot the
F5
Math
y=1/x
Y=
A: Tangent ENTER
WINDOW
2
tangent equation
ENTER
6  x  6
3  y  3
x scl for
 1 x = -2
Repeat
y scl  1
GRAPH

Review:
y
average slope: m 
x
f  a  h  f  a
slope at a point: m  lim
h 0
h
average velocity:
Vave
instantaneous velocity:
velocity = slope
total distance

total time
These are
often
mixed up
by
Calculus
students!
So are these!
If f  t  is the position function:
f t  h   f t 
V  lim
h 0
h
p