Download Math 124

This Week
Sections 2.1-2.3,2.5,2.6
First homework due Tuesday night at 11:30 p.m.
Average and instantaneous velocity worksheet Tuesday
available at http://www.math.washington.edu/∼m124/
(under week 2)
print it out before coming to class
Professor Christopher Hoffman
Math 124
Velocity
If an arrow is shot upward on the moon with a velocity of 58
meters per second. It height in meters is given by
p(t) = 58t − 0.83t 2 .
1
Find the average velocity over the time intervals
[1, 2], [1, 1.5], [1, 1.1], [1, 1.01],
2
Find the instantaneous velocity at t = 1.
Professor Christopher Hoffman
Math 124
and [1, 1.001].
Average Velocity between 1 and t
Average velocity is change in position divided by change in
time.
Average velocity between time 1 and t is
t
2
1.5
1.1
1.01
1.001
Professor Christopher Hoffman
p(t)−h(1)
t−1
55.51
55.925
56.257
56.3317
56.339
Math 124
p(t) − p(1)
.
t −1
Instantaneous Velocity
To see what happens as h approaches 0 we do a little algebra.
p(1 + h) − p(1)
(1 + h) − 1
58(1 + h) − 0.83(1 + h)2 − (58 − 0.83(1)2 )
h
58 + 58h − 0.83(1 + 2h + h2 ) − (58 − 0.83)
=
h
58h − 1.66h − 0.83h2 )
=
h
= 56.34 − 0.83h
=
As h approaches 0 the average velocity approaches 56.34.
We say the instantaneous velocity at t = 1 is 56.34 and write
Instantaneous velocity = lim
h→0
Professor Christopher Hoffman
p(1 + h) − p(1)
= 56.34.
h
Math 124
Geometric Interpretation
p(1+h)−p(1)
h
is the slope of the secant line between (1, p(1)) and
(1 + h, p(1 + h)).
As h approaches 0 the secant lines approach the tangent line
at (1, p(1)).
The instantaneous velocity of 56.34 is the slope of the tangent
line at (1, p(1)).
Professor Christopher Hoffman
Math 124
Two Interpretations
All of our work over the next few weeks can be interpreted as
finding methods to determine the instantaneous velocity for an
arbitrary position function.
Geometrically, our work can be interpreted as determining the
slope of a tangent line for an arbitrary point on the graph of an
arbitrary function.
In order to do this we first must discuss what we mean by the
limit of a function as we approach a point.
Professor Christopher Hoffman
Math 124
Limits
limx→a f (x) = L means that we can make the value of f (x)
arbitrarily close to L by taking x sufficiently close to a but not
equal to a.
In this picture limx→2 f (x) = 4.
Professor Christopher Hoffman
Math 124
limx→a f (x) does not depend on the value of f (a).
In all of these pictures limx→0 f (x) = 1.
For most of the limits limx→a f (x) = L that we take in this course
f (a) will not be defined.
Professor Christopher Hoffman
Math 124
Naive Idea
We will plug in small values of x to try to find
lim
x→0
x
±1
±1/2
±.4
±.3
±.001
sin(x)
.
x
sin(x)
x
.84147
.95885
.973545
.985067
.99999
Based on this we might be tempted to say limx→0 sin(x)/x = 1.
Professor Christopher Hoffman
Math 124
Graphical Interpretation
Looking at the graph of y = sin x/x we can see that
lim sin(x)/x = 1.
x→0
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
Math 124
Precise Definition of a Limit
The precise definition of a limit is contained in Section 2.4. We
won’t cover this section in class. We will be content to know
limits when we see them.
Professor Christopher Hoffman
Math 124
Problematic Example
Find
lim sin(π/x).
x→0
x
±1
±1/2
±1/3
±1/10
±1/100
±1/n
sin(π)
x
0
0
0
0
0
0
Based on this we might be tempted to say
lim sin(π/x) = 0.
x→0
Professor Christopher Hoffman
Math 124
But if we would have chosen
x=
2
1 + 4n
then we would have seen a different behavior as the points
approached 0.
For these values
π
+ 2nπ = 1.
sin(π/x) = sin
2
Also if we would have chosen
x=
2
3 + 4n
then we would have sin(π/x) = sin
Professor Christopher Hoffman
3π
2
Math 124
+ 2nπ = −1.
sin(π/x) oscillates between −1 and 1 infinitely often as x → 0.
Thus
lim sin(π/x)
x→0
does not exist.
Professor Christopher Hoffman
Math 124
Check your work
On every limit calculation where you want to find
lim f (x)
x→a
you should plug in values of x near a.
If the numbers you get are not close to your answer then you
did something wrong.
Professor Christopher Hoffman
Math 124
Calculating limits graphically
Professor Christopher Hoffman
Math 124
Calculating limits graphically
For the function h whose graph is given, state the value of each
quantity if it exists, If it does not exist, explain why
1
limx→−3− h(x)
2
limx→−3+ h(x)
3
limx→−3 h(x)
4
h(−3)
5
limx→0 h(x)
6
h(0)
7
limx→2 h(x)
8
h(2)
9
limx→5+ h(x)
10
limx→5− h(x)
Professor Christopher Hoffman
Math 124
Calculating limits graphically
For the function h whose graph is given, state the value of each
quantity if it exists, If it does not exist, explain why
1
limx→−3− h(x) = 4
2
limx→−3+ h(x) = 4
3
limx→−3 h(x) = 4
4
h(−3) does not exist
5
limx→0 h(x) does not exist
6
h(0) = 1
7
limx→2 h(x) = 2
8
h(2) does not exist
9
limx→5+ h(x) = 3
10
limx→5− h(x) does not exist.
Professor Christopher Hoffman
Math 124
Infinite limits
Look at a graph of the function f (x) = ln(x).
The graph has a vertical asymptote at x = 0.
As x → 0+ the function ln(x) gets more and more negative.
We say
lim+ ln(x) = −∞.
x→0
Professor Christopher Hoffman
Math 124
For every n the line x = π/2 + nπ is a vertical asymptote.
We can see that the one sided limits are
lim
x→π/2−
tan(x) = ∞
and
Professor Christopher Hoffman
lim
x→π/2+
Math 124
tan(x) = −∞.
The line x = a is a vertical asymptote of y = f (x) if at least one
of the following is true
limx→a f (x) = ∞ or −∞
limx→a+ f (x) = ∞ or −∞
limx→a− f (x) = ∞ or −∞
Professor Christopher Hoffman
Math 124
Calculating limits graphically
For the function R whose graph is given, state the following
1
limx→2 R(x)
2
limx→5 R(x)
3
limx→−3− R(x)
4
limx→−3+ R(x)
5
The equations of the vertical asymptotes.
Professor Christopher Hoffman
Math 124
Calculating limits graphically
1
limx→2 R(x) = −∞
2
limx→5 R(x) = ∞
3
limx→−3 R(x) does not exist because
limx→−3− R(x) = −∞ while limx→−3+ R(x) = ∞
4
The equations of the vertical asymptotes are x = −3,
x = 2 and x = 5.
Professor Christopher Hoffman
Math 124
A function f is continuous at a if
lim f (x) = f (a).
x→a
This definition requires three things to happen.
1
f (a) exists
2
limx→a+ f (x) = f (a)
3
limx→a− f (x) = f (a)
Professor Christopher Hoffman
Math 124
Graphs of Continuous Functions
This function is continuous at every point except x = −2, 2, 4
and 6.
Professor Christopher Hoffman
Math 124
Continuous Functions
Any function defined by the following functions is continuous at
every point in its domain.
polynomials
root functions
trigonometric functions
inverse trigonometric functions
exponential functions and
logarithmic functions
Professor Christopher Hoffman
Math 124
Examples of Continuous Functions
All of the following functions are continuous wherever they are
defined.
f (x) = sin
3+x
x 2 +1
g(t) = e3t/4+ln(2t)
√
x 3 +1
r (x) = 2x−5
√
m(t) = 3t + π tan−1 (4 cos(t − π))
Professor Christopher Hoffman
Math 124
Where are the following functions continuous
1
2
3
x 2 sin(x)
(x−3)(x+4) is continuous except when x = 3
ln(x 2 − 1) is continuous except when −1 ≤
ex/(x−1)
is continuous except when x = 1.
Professor Christopher Hoffman
Math 124
and x = −4
x ≤1
Evaluating Limits of Continuous Functions
This is the easiest thing we will do all class all quarter.
If f (x) is continuous at a then
lim f (x) = f (a).
x→a
Professor Christopher Hoffman
Math 124
Evaluating Limits of Continuous Functions
Find the following limits.
√
1
limx→3
2
limy →−2
3
limx→−2
3πx+1
x 2 −6
sin
√
√
√
=
3y +10
5y 2 −3
3cx+12
7cx+1
=
9π+1
3
=
√
sin(2)
17
12−6cx
1−14c
for any c < 2 and c 6= 1/14.
When evaluating limits in this course evaluating the function at
the point should always be your first approach.
If this fails then try pugging in numbers close to a. After you
make a guess you must justify your answer.
Professor Christopher Hoffman
Math 124
Find
(2 + h)3 − 8
.
h
h→0
lim
Plugging in h = 0 we get 00 .
(2 + h)3 − 8
h
h→0
lim
(2 + h)3 − 8
h
h→0
8 + 12h + 6h2 + h3 − 8
= lim
h
h→0
12h + 6h2 + h3
= lim
h
h→0
= lim 12 + 6h + h2
=
lim
h→0
= 12.
If two functions agree everywhere except at one point a then
the limits approaching a are the same.
Professor Christopher Hoffman
Math 124
Find
1/(7 + 3t) − 1/7
.
t
t→0
lim
Plugging in t = 0 we get 00 .
1/(7 + 3t) − 1/7
t
t→0
lim
Professor Christopher Hoffman
7 − (7 + 3t)
t→0 t(7 + 3t)(7)
−3t
= lim
t→0 t(7 + 3t)7
−3
= lim
t→0 (7 + 3t)7
3
= − .
49
=
lim
Math 124
Rationalize the denominator
Find
16 − t
√ .
t→16 4 − t
lim
Plugging in t = 16 we get 00 .
16 − t
√
lim
t→16 4 − t
√
(16 − t)(4 + t)
√
√
= lim
t→16 (4 − t)(4 + t)
√
(16 − t)(4 + t)
= lim
(16 − t)
t→16
√
= lim 4 + t
t→16
= 8
Professor Christopher Hoffman
Math 124
Find
lim
x→0
3
3
− 3
x
x +x
.
Plugging in x = 0 we get ∞ − ∞.
lim
x→0
3
3
− 3
x
x +x
=
=
=
=
=
Professor Christopher Hoffman
3(x 2 + 1) − 3
x→0
x3 + x
3x 2
lim 3
x→0 x + x
3x
lim 2
x→0 x + 1
3·0
02 + 1
0.
lim
Math 124
Find
x 3 − 27
x→3 x − 3
lim
Plugging in x = 3 we get 00 .
x 3 − 27
x→3 x − 3
lim
(x − 3)(x 2 + 3x + 9)
x −3
x→0
2
= lim x + 3x + 9
=
lim
x→0
= 27.
Professor Christopher Hoffman
Math 124
Limit Laws
Suppose c is a constant and the limits
lim f (x)
x→a
and
lim g(x)
x→a
exist. Then
1
2
3
4
limx→a f (x) + g(x) = limx→a f (x) + limx→a g(x)
limx→a f (x) − g(x) = limx→a f (x) − limx→a g(x)
limx→a cf (x) = c limx→a f (x)
limx→a f (x) · g(x) = limx→a f (x) · limx→a g(x)
5
f (x)
limx→a f (x)
=
x→a g(x)
limx→a g(x)
lim
if limx→a g(x) 6= 0
Professor Christopher Hoffman
Math 124
Find
√
3π
lim cos(
) − 3 (3(t + 1)2 −4t).
t +1
t→−1
This looks like a mess. Let’s look at the graph.
Professor Christopher Hoffman
Math 124
Find
√
3π
lim cos(
) − 3 (3(t + 1)2 −4t).
t +1
t→−1
This looks like a mess. Let’s look at the graph.
Professor Christopher Hoffman
Math 124
Here is the graph of the function
√
3π
g(t) = cos(
) − 3 (3(t + 1)2 −4t)
t +1
along with
√
f (t) = −2(3(t + 1)2 −4t)
and
√
h(t) = −4(3(t + 1)2 −4t).
Notice that as t → −1 all three functions are approaching 0.
Professor Christopher Hoffman
Math 124
As these are continuous functions and defined at t = −1 we get
p
√
lim −2(3(t + 1)2 −4t) = −2(3(−1 + 1) −4(−1) = 0
t→−1
and
p
√
lim −4(3(t + 1)2 −4t) = −4(3(−1 + 1) −4(−1) = 0.
t→−1
The function that we are interested in
√
3π
) − 3 (3(t + 1)2 −4t)
g(t) = cos(
t +1
is sandwiched in between so limt→−1 g(t) must be zero as well.
Professor Christopher Hoffman
Math 124
Sandwich Theorem
This theorem makes the previous slide precise.
Theorem
If f (x) ≤ g(x) ≤ h(x) in some interval around a and
lim (f (x)) = lim (h(x)) = L
x→a
x→a
then
lim (g(x)) = L
x→a
Professor Christopher Hoffman
Math 124
Find
x − 14
lim √
.
x→14
14 + x
First we plug in x = 14 and get
14 − 14
0
√
= √ = 0.
14 + 14
28
As this is a continuous function defined at 14 we have
x − 14
= 0.
lim √
x→14
14 + x
Professor Christopher Hoffman
Math 124
Limit Laws and Graphs
Here is the graph of a function f (x)
Find the following limits.
1
limx→2 (f (x)2 )
2
limx→−2 (f (x)f (x + 3))
3
limx→5 f (x)(2 − f (x − 3))2
Professor Christopher Hoffman
Math 124
Let f (x) =
x−1
.
(x+2)x 2
1
Find all vertical asymptotes of f (x).
2
Find the one sided limits of f (x) as x approaches the
asymptotes.
The function is a rational function so it is continuous at every
point where it is defined.
It is defined everywhere the denominator is not zero, which is
when x = 0 and x = −2.
It is important to keep signs right. We must be careful about the
terms close to zero.
Professor Christopher Hoffman
Math 124
As x → 0+ , x 2 approaches 0 from the right. Also when x → 0− ,
x 2 approaches 0 from the right.
As x → 0 the numerator x − 1 approaches −1 and x + 2
approaches 2.
Thus as x → 0 from either side the numerator is negative and
the denominator is small and positive. The fraction is thus large
and negative. Because of this
lim
x→0
x −1
= −∞.
(x + 2)x 2
Professor Christopher Hoffman
Math 124
As x → −2+ the numerator x − 1 approaches −3 and the
denominator approaches 0 from the positive side.
As the numerator is negative and the denominator is small and
positive the fraction is large and negative. Because of this
x −1
lim +
= −∞.
x→−2 (x + 2)x 2
As x → −2− the numerator x − 1 approaches −3 and the
denominator approaches 0 from the negative side.
As the numerator is negative and the denominator is small and
negative the fraction is large and positive. Thus
x −1
lim
= ∞.
−
x→−2 (x + 2)x 2
Professor Christopher Hoffman
Math 124
When we graph the function we can see that there are vertical
asymptotes at x = −2 and x = 0.
Professor Christopher Hoffman
Math 124
x−1
.
(x+2)x 2
1
Find all vertical asymptotes of
2
Find the one sided limits of f (x) as x approaches the
asymptotes.
The function is a rational function so it is continuous at every
point where it is defined.
It is defined everywhere the denominator is not zero, which is
when x = 0 and x = −2.
It is important to keep signs right. We must be careful about the
terms close to zero.
Professor Christopher Hoffman
Math 124
As x → 0+ , x 2 approaches 0 from the right. Also when x → 0− ,
x 2 approaches 0 from the right.
As x → 0 the numerator x − 1 approaches −1 and x + 2
approaches 2.
Thus as x → 0 from either side the numerator is negative and
the denominator is small and positive. The fraction is thus large
and negative. Because of this
lim
x→0
x −1
= −∞.
(x + 2)x 2
Professor Christopher Hoffman
Math 124
As x → −2+ the numerator x − 1 approaches −3 and the
denominator approaches 0 from the positive side.
As the numerator is negative and the denominator is small and
positive the fraction is large and negative. Because of this
x −1
lim +
= −∞.
x→−2 (x + 2)x 2
As x → −2− the numerator x − 1 approaches −3 and the
denominator approaches 0 from the negative side.
As the numerator is negative and the denominator is small and
negative the fraction is large and positive. Thus
x −1
lim
= ∞.
−
x→−2 (x + 2)x 2
Professor Christopher Hoffman
Math 124
Limit Laws
Suppose c is a constant and the limits
lim f (x)
x→a
and
lim g(x)
x→a
exist. Then
1
2
3
4
5
limx→a f (x) + g(x) = limx→a f (x) + limx→a g(x)
limx→a f (x) − g(x) = limx→a f (x) − limx→a g(x)
limx→a cf (x) = c limx→a f (x)
limx→a f (x) · g(x) = limx→a f (x) · limx→a g(x)
limx→a
f (x)
g(x)
=
limx→a f (x)
limx→a g(x)
if limx→a g(x) 6= 0
Professor Christopher Hoffman
Math 124
To infinity and beyond
Professor Christopher Hoffman
Math 124
Definition of Limits at ∞
We write
lim f (x) = L
x→∞
when the values of f (x) can be made arbitrarily close to L by
taking x sufficiently large. Similarly we write
lim f (x) = L
x→−∞
when the values of f (x) can be made arbitrarily close to L by
taking sufficiently large negative values for x.
Professor Christopher Hoffman
Math 124
Limits at ∞ Graphically
Consider the function f (x) = 3 − 2e−x
As x gets large f (x) gets closer to 3. The graph of f (x) gets
closer to the graph of y = 3 and we write
lim 3 − 2e−x = 3.
x→∞
Professor Christopher Hoffman
Math 124
Limits at ∞ Graphically
As x gets more and more negative 3 − 2e−x gets more and
more negative as well. We write
lim 3 − 2e−x = −∞.
x→−∞
Professor Christopher Hoffman
Math 124
Horizontal Asymptotes
The line y = L is a horizontal asymptote of the curve y = f (x) if
either
lim f (x) = L
x→∞
or
lim f (x) = L.
x→−∞
Thus y = π/2 and y = −π/2 are horizontal asymptotes and
lim tan−1 (x) = π/2
x→−∞
and
Professor Christopher Hoffman
lim tan−1 (x) = −π/2
x→∞
Math 124
Limits of polynomials at ∞
If r > 0 then limx→∞ x r = ∞
If r < 0 then limx→∞ x r = 0
If r is even then limx→−∞ x r = ∞
If r is odd then limx→−∞ x r = −∞
Professor Christopher Hoffman
Math 124
Limits Laws at ∞
Suppose c is a constant and the limits
lim f (x)
x→∞
and
lim g(x)
x→∞
exist. Then
1
2
3
4
5
limx→∞ f (x) + g(x) = limx→∞ f (x) + limx→∞ g(x)
limx→∞ f (x) − g(x) = limx→∞ f (x) − limx→∞ g(x)
limx→∞ cf (x) = c limx→∞ f (x)
limx→∞ f (x) · g(x) = limx→∞ f (x) · limx→∞ g(x)
limx→∞
f (x)
g(x)
=
limx→∞ f (x)
limx→∞ g(x)
Professor Christopher Hoffman
if limx→∞ g(x) 6= 0
Math 124
Extended Real Numbers
When working with limits it is useful to use the extended real
numbers.
∞·∞=∞
∞+∞=∞
∞+a=∞
a∞ = ∞ if a > 0
a∞ = −∞ if a < 0
∞ − ∞ =???
Professor Christopher Hoffman
Math 124
Rational Functions
Find the following limits.
limx→∞ (−5x 3 + 2(r − 1)x)
x 3/2 +2πx 7 −2x
1
−5x 3 +2(r −1)x
√
limx→−∞
x 4 +1
limx→∞
Professor Christopher Hoffman
Math 124
Polynomials
With polynomials the largest term dominates. To see this we
factor out the largest power of x.
lim (−5x 3 + 2(r − 1)x) =
x→∞
=
lim x 3 (−5 + 2(r − 1)/x 2 )
x→−∞
lim x 3 lim (−5 + 2(r − 1)/x 2 )
x→−∞
x→−∞
= (−∞)(−5)
= ∞
Professor Christopher Hoffman
Math 124
Divide by the largest power of the denominator
For functions like this we divide top and bottom by the largest
power of the denominator.
x 3/2 + 2πx 7 − 2x
x→∞
x2 − 1
lim
=
(1/x 2 )(x 3/2 + 2πx 7 − 2x)
x→∞
(1/x 2 )(x 2 − 1)
=
x −1/2 + 2πx 5 − 2/x)
x→∞
1 − 1/x 2
lim
lim
limx→∞ (x −1/2 + 2πx 5 − 2/x)
limx→∞ (1 − 1/x 2 )
0+∞+0
=
1−0
= ∞
=
Professor Christopher Hoffman
Math 124
Divide by the largest power of the denominator
For this problem we want to divide by x 2 =
−5x 3 + 2(r − 1)x
√
x→−∞
x4 + 1
lim
=
=
=
=
=
Professor Christopher Hoffman
√
x 4.
(1/x 2 )(−5x 3 + 2(r − 1)x)
√
x→−∞
(1/x 2 ) x 4 + 1
−5x + 2(r − 1)/x
p
lim
x→−∞
1 + 1/x 4
limx→−∞ (−5x + 2(r − 1)/x)
p
limx→−∞ 1 + 1/x 4
−5(−∞) + 0
1
−∞
lim
Math 124
lim
x→∞
p
x 2 + 3x + 2 − x
We rationalize the numerator by multiplying the expression by
√
x 2 + 3x + 2 + x
√
x 2 + 3x + 2 + x
p
2
lim
x + 3x + 2 − x
x→∞
p
√x 2 + 3x + 2 + x
2
= lim
x + 3x + 2 − x √
x→∞
x 2 + 3x + 2 + x
x 2 + 3x + 2 − x 2
= lim √
x→∞
x 2 + 3x + 2 + x
3x + 2
= lim √
2
x→∞
x + 3x + 2 + x
Professor Christopher Hoffman
Math 124
Then we multiply numerator and denominator by 1/x.
lim
x→∞
p
x 2 + 3x + 2 − x
=
=
=
=
=
Professor Christopher Hoffman
(3x + 2)(1/x)
√
2
x→∞ ( x + 3x + 2 + x)(1/x)
3 + 2/x
lim p
x→∞ ( 1 + 3/x + 2/x 2 + 1)
limx→∞ (3 + 2/x)
p
limx→∞ ( 1 + 3/x + 2/x 2 + 1)
3
1+1
3
2
lim
Math 124
For all real numbers a > 0, b and c find
p
p
ax 2 + bx + c − 5x
and
lim
ax 2 + bx + c − 5x
lim
x→∞
x→−∞
As in the previous problem we will multiply the expression times
√
ax 2 + bx + c + 5x
√
ax 2 + bx + c + 5x
and then factor out the largest power of the denominator.
Professor Christopher Hoffman
Math 124
lim
x→∞
=
=
=
=
=
=
=
p
ax 2 + bx + c − 5x
√
p
ax 2 + bx + c + 5x
lim ( ax 2 + bx + c − 5x) √
x→∞
ax 2 + bx + c + 5x
2
2
ax + bx + c − 25x
lim √
x→∞
ax 2 + bx + c + 5x
(a − 25)x 2 + bx + c
lim √
x→∞
ax 2 + bx + c + 5x
(1/x)(a − 25)x 2 + bx + c)
√
lim
x→∞ (1/x)( ax 2 + bx + c + 5x)
(a − 25)x + b + c/x
lim p
x→∞
a + b/x + c/x 2 + 5
limx→∞ ((a − 25)x + b + c/x)
p
limx→∞ ( a + b/x + c/x 2 + 5)
limx→∞ ((a − 25)x + b + 0)
√
a
Professor Christopher Hoffman
Math 124
p
lim
x→∞ ((a − 25)x + b + 0)
√
ax 2 + bx + c − 5x =
x→∞
a
lim
If a > 25 then the leading term has a positive coefficient and
the limit approaches ∞.
If a < 25 then the leading term has a negative coefficient and
the limit approaches ∞.
If a = 25 then the limit becomes b/5.
The limits as x → −∞ are found the the same way.
Professor Christopher Hoffman
Math 124
Find
lim
x→∞
p
4x 4 + 5x 3 + 10 − (2x 2 + x) .
Again we will multiply by the conjugate and divide by the
highest power of x in the denominator.
p
lim
4x 4 + 5x 3 + 10 − (2x 2 + x)
x→∞
√4x 4 + 5x 3 + 10 + (2x 2 +
p
2
4
3
4x + 5x + 10 − (2x + x) √
= lim
x→∞
4x 4 + 5x 3 + 10 + (2x 2 +
(4x 4 + 5x 3 + 10) − (2x 2 + x)2
= lim √
x→∞
4x 4 + 5x 3 + 10 + (2x 2 + x)
(4x 4 + 5x 3 + 10) − (4x 4 + 4x 3 + x 2 )
√
= lim
x→∞
4x 4 + 5x 3 + 10 + (2x 2 + x)
x 3 − x 2 + 10
= lim √
x→∞
4x 4 + 5x 3 + 10 + (2x 2 + x)
Professor Christopher Hoffman
Math 124
x − 1 + 10/x 2
x→∞
2
= ∞
=
lim
q
4 + 5/x + 10/x 4 + 2 + 1/x 2
Professor Christopher Hoffman
Math 124
x−1
.
(x+2)x 2
x−1
of (x+2)x
2.
1
Find all vertical asymptotes of
2
Find all horizontal asymptotes
3
Find the one sided limits of f (x) as x approaches the
asymptotes.
The function is a rational function so it is continuous at every
point where it is defined.
It is defined everywhere the denominator is not zero, which is
when x = 0 and x = −2.
It is important to keep signs right. We must be careful about the
terms close to zero.
Professor Christopher Hoffman
Math 124
As x → 0+ , x 2 approaches 0 from the right. Also when x → 0− ,
x 2 approaches 0 from the right.
As x → 0 the numerator x − 1 approaches −1 and x + 2
approaches 2.
Thus as x → 0 from either side the numerator is negative and
the denominator is small and positive. The fraction is thus large
and negative. Because of this
lim
x→0
x −1
= −∞.
(x + 2)x 2
Professor Christopher Hoffman
Math 124
As x → −2+ the numerator x − 1 approaches −3 and the
denominator approaches 0 from the positive side.
As the numerator is negative and the denominator is small and
positive the fraction is large and negative. Because of this
x −1
lim +
= −∞.
x→−2 (x + 2)x 2
As x → −2− the numerator x − 1 approaches −3 and the
denominator approaches 0 from the negative side.
As the numerator is negative and the denominator is small and
negative the fraction is large and positive. Thus
x −1
lim
= ∞.
−
x→−2 (x + 2)x 2
Professor Christopher Hoffman
Math 124
Sketch a graph of a function f (x) with the following properties.
1
has horizontal asymptotes y = 0 and y = 4
2
has vertical asymptotes x = −3 and x = 4
3
limx→4 f (x) = ∞
4
limx→−3 f (x) does not exist.
Professor Christopher Hoffman
Math 124
Graphical limits
Composition of functions.
Change of variables.
Professor Christopher Hoffman
Math 124