CSC Worthy
Warm-up
Evaluate. Use your calculator and round to the
nearest hundredth.
1. sin32.5 = cos _______
= .54
57.5
2. csc12.06
= sec _______
77.94 = 4.79
3. log6127 = 2.70
Warm-up
(
)
1. sin 5 3 = − 3 2
3. cos
5. 2
−1
(
2log 2 8
− 2
= 64
)
(
)
2. tan − 3 4 = 1
−1
3

= 4 4. tan ( 0 ) = 0
2
6. e
−2ln 7
= 149
7. 3§4.2¨ − 4§−2.5¨ + 2§−5¨ = 14
2
1
−
x
−1
8. tan ( cos x ) = x
CSC Worthy
Warm-up
Sketch the graph of a function that satisfies all
of the following properties.
f (0) = −1, f (3) = 1,lim
f
(
x
)
=
1,lim
f
(
x
)
=
−
2,lim
f
(
x
)
=
2
x →0
x→3−
x→3+
Warm-up
CSC Worthy
Section 2.1:5
If a ball is thrown into the air with a velocity of
40ft/s, the height in feet t seconds later is given
by y = 40t – 16t2.
a) Find the average velocity for the time
period beginning when t = 2 and lasting . . .
(i). O.5 seconds (ii). 0.1 seconds
(iii). 0.05 seconds (iv). 0.01 seconds
b) Estimate the instantaneous velocity when
t = 2.
Warm-up
f ( x + h) − f ( x )
m=
h
CSC Worthy
Section 2.1:5
If a ball is thrown into the air with a velocity of
40ft/s, the height in feet t seconds later is given
by y = 40t – 16t2.
a) Find the average velocity for the time
period beginning when t = 2 and lasting . . .
(i). O.5 seconds (ii). 0.1 seconds
y ( 2 + h ) − y (2)
y ( 2.5 ) − y (2)
vavg . =
vavg . =
(2 + h) − 2
( 2.5) − 2
or
0 − 16
=
= −32ft/s
= −24 − 16h, if h  0
.5
Warm-up
f ( x + h) − f ( x )
m=
h
CSC Worthy
Section 2.1:5
If a ball is thrown into the air with a velocity of
40ft/s, the height in feet t seconds later is given
by y = 40t – 16t2.
a) Find the average velocity for the time
period beginning when t = 2 and lasting . . .
(i). O.5 seconds
(ii).
0.1
seconds
−32ft/s
−25.6ft/s
Is the ball
rising or
falling?
Warm-up
f ( x + h) − f ( x )
m=
h
CSC Worthy
Section 2.1:5
If a ball is thrown into the air with a velocity of
40ft/s, the height in feet t seconds later is given
by y = 40t – 16t2.
a) Find the average velocity for the time
period beginning when t = 2 and lasting . . .
(i). O.5 seconds
(ii).
0.1
seconds
−32ft/s
−25.6ft/s
(iii). 0.05 seconds (iv). 0.01 seconds
−24.2ft/s
−24.8ft/s
b) Estimate the instantaneous velocity when
t = 2. The instantaneous velocity when t = 2 is 24ft/s.
CSC Worthy
Warm-up
Fill in the table and use it to estimate . . .
ln x − ln 4
lim
x →4
x−4
Record decimals to 6 places.
.251576
.250313
.250031
.25
.249967
.249688
.248450
Warm-up
Use the graph below to identify all points at which f is
not continuous and state the type of discontinuity.
For each of these points, then determine if it is
continuous from the left, the right, or neither.
x=
Type of
Discontinuity
Continuous
from . . .
-5
removable
neither
-3
jump
left
1
removable
neither
5
infinite
neither
−5 −3
1
5
MA121 Calculus
2.6: Limits at Infinity
(Horizontal Asymptotes)
Dr. Soltys
+
Limits at Infinity
Precise Definition of a Limit at Infinity
 Let f be a function defined on some
interval (a, ). Then . . .
lim f ( x ) = L
means that for every   0 there is a
x →
corresponding number N such that
if x  N then f x − L   .
( )
Intuitively, this means that the values of f(x) can be made
arbitrarily close to L by requiring x to be sufficiently large.
+
Limits at Infinity
Precise Definition of a Limit at Infinity
lim f ( x ) = L
if
xN
x →
then
f ( x) − L   .
−
Limits at Infinity
Precise Definition of a Limit at Infinity
 Let f be a function defined on some
interval (- ,a). Then . . .
lim f ( x ) = L
means that for every   0 there is a
x →−
corresponding number N such that
if x  N then f x − L   .
( )
Intuitively, this means that the values of f(x) can be made
arbitrarily close to L by requiring x to be sufficiently small.
−
Limits at Infinity
Precise Definition of a Limit at Infinity
lim f ( x ) = L
if
xN
x →−
then
f ( x) − L   .
Limits at Infinity
Horizontal Asymptotes
The line y = L is called a horizontal asymptote
of the curve y = f(x) iff . . .
lim f ( x ) = L
x →
or
lim f ( x ) = L
x →−
Horizontal asymptotes are always of the form
y=L
Limits at Infinity
Vertical Asymptotes
Horizontal Asymptotes
lim f ( x) = 
lim f ( x) = b
x→a
Equation: x = a
x=3
lim f ( x) = −
x →3−
lim f ( x) = 
x →3+
x → 
Equation: x = b
Limits at Infinity
Vertical Asymptotes
Horizontal Asymptotes
lim f ( x) = 
lim f ( x) = b
x→a
Equation: x = a
lim f ( x) = −
x →0 −
x=0
lim f ( x) = −
x →0 +
x → 
Equation: x = b
Limits at Infinity
Vertical Asymptotes
Horizontal Asymptotes
lim f ( x) = 
lim f ( x) = b
x → 
x→a
Equation: x = a
Equation: y = b
lim f ( x) = 1
x →−
lim f ( x) = −
x →0 −
x=0
lim f ( x) = −
x →0 +
2
lim f ( x) = 1
x →
y= 1
2
2
Limits at Infinity
Here are some common functions with limits
at infinity that you should be able to quickly
recall and sketch.
Limits at Infinity
1. Exponential and Logarithmic Functions
y=0
x
lim
e
=0
x →−
−x
lim
2
=
x →−
x
lim
e
=
x →
−x
lim
2
=0
x →
y=0
lim(− log 2 x) = 
lim ln x = −
x →0 +
x=0
lim
ln x = 
x →
x →0 +
x=0
lim(
− log 2 x) = 
x →
Limits at Infinity
2. Arctangent
y =
−1

lim
tan
x
=
x →
2
2
y = −
−1

lim
tan
x
=
−
x →−
2
2
Limits at Infinity
3. Graphs of
y=0
1
x
Horizontal Asymptotes?
Vertical Asymptotes?
n
lim 1 = 0
x →−
x
1
lim
x →−
x2
1 =0
lim
x →
x
1
lim
x →
x2
y=0
x=0
=0
x=0
y=0
1
lim
x →−
lim 1
x →
x=0
=0
x
3
x3
=0
=0
1 =0
lim
−
x →−
x
lim
− 1 =0
x →
x
y=0
x=0
Limits at Infinity
4. Polynomial Functions of Degree n
Even-degree Polynomial
Odd-degree Polynomial
+ leading
coefficient
lim f ( x) = 
lim f ( x) = −
x →−
x →−
lim
f ( x) = 
x →
lim
f ( x) = 
x →
End Behavior: What happens to f(x) as x approaches positive
and negative infinity?
Limits at Infinity
4. Polynomial Functions of Degree n
Even-degree Polynomial
f ( x) = x 4 − 2 x 3 + 2 x 2 − 1
Odd-degree Polynomial
What if the
leading
coefficient is - ?
f ( x) = .06 x 3 + .05 x + 1
f ( x) = x 4 − 2 x 2 − 1
ff((xx))==−xx3 3−+33xx2 2+−11
f (f x()x=
) =− x 4 −
+ 2 x3 +
− 2x2 +
−1
f ( x) = x 5 − 4 x 4 − 3x 3 − x 2 − 3x − 5
lim f ( x) = 
x →
lim f ( x) = − and lim
f ( x) = 
x →
x →−
How to Calculate Limits at Infinity for
Rational Functions
Horizontal Asymptotes
7 x + 16 x + 2 x
How do I find lim
?
3
x→
13 x − 2 x − 32
3


2
Horizontal asymptotes can be found by finding the
limit of the quotient when the numerator and
denominator are divided by the greatest power of x as
x approaches  
If the rational function is of the form polynomial/polynomial,
then . . .
P( x)
P( x)
lim
x →−
Q( x)
= lim
x →
Q( x)
and there is only one horizontal asymptote.
7 x + 16 x + 2 x
Find the horizontal asymptote for f ( x) =
13 x 3 − 2 x − 32
3
To find the limit as x
Divide the numerator
and the denominator by
the largest power of x
...
3
2
7 x 16 x 2 x
+ 3 + 3
3
x
f ( x) = x 3 x
13x 2 x 32
− 3− 3
3
x
x
x
2
Verify with table.
16 2
+ 2
x x
Simplify f ( x) =
Therefore, y = 713
2 32
13 − 2 − 3
is a horizontal
x
x
asymptote for f(x)
16 2
7+ + 2
7+0+0
x
x
=
Find the limit of f(x) as lim
2 32
x →
13 − 2 − 3 13 − 0 − 0
x approaches
x
x
7+
How to Calculate Limits at Infinity for
Do you notice any
Rational Functions
pattern? When will
Horizontal Asymptotes
we get y =
4x3 + 2x 2 + 5
 f ( x) = 3
8 x − 7 x − 12
Horizontal: y = 1
2
Is this possible?
6 x 2 − 13x + 8
 g ( x) =
5 x 2 + 11x
Horizontal: y = 6
12 x 3 − x 2 + x − 23
 h( x ) =
21x 4 − 15 x 2 + 2 x
Horizontal: y = 0
x 3 + 8 x − 13
 f ( x) = 2
9 x + 3x − 54
5
no horizontal asymptotes
This graph appears to be asymptotic to a “diagonal” line.
This line is a slant or oblique asymptote.
0 or ?
How to Calculate Limits at Infinity for
Rational Functions
Horizontal Asymptotes
Horizontal Asymptote “Shortcut”



If the degree of the numerator and the denominator are
equal, then the horizontal asymptote is the ratio of the
leading coefficients
If the degree of the numerator is less than the degree of
the denominator, then the horizontal asymptote is y = 0
If the degree of the numerator is greater than the
degree of the denominator, then there is no horizontal
asymptote
How to Calculate Limits at Infinity for
Do you notice any
Rational Functions
pattern? When will
Horizontal Asymptotes
Verify the shortcuts.
4x3 + 2x 2 + 5
 f ( x) = 3
8 x − 7 x − 12
Horizontal: y = 1
6 x 2 − 13x + 8
 g ( x) =
5 x 2 + 11x
Horizontal: y = 6
12 x 3 − x 2 + x − 23
 h( x ) =
21x 4 − 15 x 2 + 2 x
Horizontal: y = 0
x 3 + 8 x − 13
 f ( x) = 2
9 x + 3x − 54
we get y =
2
5
no horizontal asymptotes
0 or ?
9x + 6
Find the horizontal asymptote for f ( x) =
5x − 1
2
To find the limit as x + . . .
Divide the numerator
and the denominator by
the largest power of x
What is does this equal
as we approach + ?
x2 = x
Simplify
Find the limit of f(x) as
x approaches
x = x2
9 x2 6
+ 2
2
x
f ( x) = x
5x
1
−
x2
x2
6
9+ 2
x
f ( x) =
5x 1
−
x x
3
9+0
=
lim
x→ 5 − 0
5
Therefore, y = 3 5
is a horizontal
asymptote for f(x)
9x + 6
Find the horizontal asymptote for f ( x) =
5x − 1
2
To find the limit as x - . . .
Divide the numerator
and the denominator by
the largest power of x
What is does this equal
as we approach - ?
x2 = x
Simplify
Find the limit of f(x) as
x approaches
9 x2 6
+ 2
2
x
f ( x) = x
5x
1
−
x2
x2
6
9+ 2
x
f ( x) =
5x 1
−
−x −x
3
9+0
=
−
lim
x→ −5 + 0
5
y=3
5
y=3
5
Therefore, y = − 3 5
is a horizontal
asymptote for f(x)
Limits at Infinity
Horizontal Asymptotes
A function may have two horizontal asymptotes
if the function has a radical variable expression
or an absolute value quantity.
Arctan also has two horizontal asymptotes.
It is incorrect to think that
lim ( x − x ) =  −  = 0
2
x →
Infinity is not a number and the
limit laws do not apply in this
situation.
How can we find this limit?
Limits at Infinity
Method 1
lim ( x − x ) = lim x ( x − 1)
2
x →
x →
= lim x  lim ( x − 1) =    = 
x →
x →
Method 2
Consider the end behavior of
a polynomial function with
degree 2.
Limits at Infinity
Vertical Asymptotes
1. If possible, factor the
numerator and
denominator.
2. Cancel any common
factors (holes).
3. Any remaining domain
restrictions are vertical
asymptotes. Set equal to
0 and solve for x.
4. State the equations of
all vertical asymptotes
as x = a.
Horizontal Asymptotes
1. Find the limit of f(x) as x
approaches positive and
negative infinity.
2. State the equations of all
horizontal asymptotes as
y = b.
Practice
Find all of the vertical
and horizontal
asymptotes for . . .
Hole:
( −2, 213)
x + 2x
y=
x − 9 x − 22
2
2
x + 2x
x
lim
= lim
=
x − 9 x − 22
x − 11
x + 2x
x
lim
= lim
= −
x − 9 x − 22
x − 11
x + 2x
lim
=1
x − 9 x − 22
x + 2x
lim
=1
x − 9 x − 22
2
x →11+
x →
2
2
Vertical :
Asymptote
Horizontal :
Asymptote
x = 11
x →11−
x →
2
2
y =1
x → +
2
2
x → −
2
Practice
Find all of the vertical
and horizontal
asymptotes for . . .
Does this
make sense?
Vertical :
Asymptote
x = −4
x=4
Horizontal :
Asymptote
y=0
2 x + 10
y=
x − 16
2
2 x + 10
lim
=
x − 16
x →−4−
2
2 x + 10
lim
= −
x − 16
x →−4+
2
2 x + 10
2 x + 10
lim
= − lim
=
x − 16
x − 16
x →4−
2
2 x + 10
lim
=0
x − 16
x →−
2
x →4+
2
2 x + 10
lim
=0
x − 16
x →
2