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This Week
Sections 2.1-2.3,2.5,2.6First homework due Tuesday night at 11:30 p.m.Average and instantaneous velocity worksheet Tuesdayavailable 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 58meters per second. It height in meters is given by
p(t) = 58t − 0.83t2.
1 Find the average velocity over the time intervals
[1,2], [1,1.5], [1,1.1], [1,1.01], and [1,1.001].
2 Find the instantaneous velocity at t = 1.
Professor Christopher Hoffman Math 124
Average Velocity between 1 and t
Average velocity is change in position divided by change intime.
Average velocity between time 1 and t isp(t)− p(1)
t − 1.
t p(t)−h(1)t−1
2 55.511.5 55.9251.1 56.257
1.01 56.33171.001 56.339
Professor Christopher Hoffman Math 124
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 = limh→0
p(1 + h)− p(1)h
= 56.34.
Professor Christopher Hoffman 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 lineat (1,p(1)).
The instantaneous velocity of 56.34 is the slope of the tangentline at (1,p(1)).
Professor Christopher Hoffman Math 124
Two Interpretations
All of our work over the next few weeks can be interpreted asfinding methods to determine the instantaneous velocity for anarbitrary position function.
Geometrically, our work can be interpreted as determining theslope of a tangent line for an arbitrary point on the graph of anarbitrary function.
In order to do this we first must discuss what we mean by thelimit 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 notequal 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 coursef (a) will not be defined.
Professor Christopher Hoffman Math 124
Naive Idea
We will plug in small values of x to try to find
limx→0
sin(x)x
.
x sin(x)x
±1 .84147±1/2 .95885±.4 .973545±.3 .985067±.001 .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
limx→0
sin(x)/x = 1.
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. Wewon’t cover this section in class. We will be content to knowlimits when we see them.
Professor Christopher Hoffman Math 124
Problematic Example
Findlimx→0
sin(π/x).
x sin(π)x
±1 0±1/2 0±1/3 0±1/10 0±1/100 0±1/n 0
Based on this we might be tempted to say
limx→0
sin(π/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 pointsapproached 0.For these values
sin(π/x) = sin(π
2+ 2nπ
)= 1.
Also if we would have chosen
x =2
3 + 4n
then we would have sin(π/x) = sin(3π
2 + 2nπ)= −1.
Professor Christopher Hoffman Math 124
sin(π/x) oscillates between −1 and 1 infinitely often as x → 0.Thus
limx→0
sin(π/x)
does not exist.
Professor Christopher Hoffman Math 124
Check your work
On every limit calculation where you want to find
limx→a
f (x)
you should plug in values of x near a.
If the numbers you get are not close to your answer then youdid 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 eachquantity 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 eachquantity if it exists, If it does not exist, explain why
1 limx→−3− h(x) = 42 limx→−3+ h(x) = 43 limx→−3 h(x) = 44 h(−3) does not exist5 limx→0 h(x) does not exist6 h(0) = 17 limx→2 h(x) = 28 h(2) does not exist9 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
limx→0+
ln(x) = −∞.
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
limx→π/2−
tan(x) =∞ and limx→π/2+
tan(x) = −∞.
Professor Christopher Hoffman Math 124
The line x = a is a vertical asymptote of y = f (x) if at least oneof 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 following1 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
limx→a
f (x) = f (a).
This definition requires three things to happen.
1 f (a) exists2 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 = −2,2,4and 6.
Professor Christopher Hoffman Math 124
Continuous Functions
Any function defined by the following functions is continuous atevery point in its domain.
polynomialsroot functionstrigonometric functionsinverse trigonometric functionsexponential functions andlogarithmic functions
Professor Christopher Hoffman Math 124
Examples of Continuous Functions
All of the following functions are continuous wherever they aredefined.
f (x) = sin(
3+xx2+1
)g(t) = e3t/4+ln(2t)
r(x) =√
x3+12x−5
m(t) =√
3t + π tan−1(4 cos(t − π))
Professor Christopher Hoffman Math 124
Where are the following functions continuous1 x2 sin(x)
(x−3)(x+4) is continuous except when x = 3 and x = −4
2 ln(x2 − 1) is continuous except when −1 ≤ x ≤ 13 ex/(x−1) is continuous except when x = 1.
Professor Christopher Hoffman Math 124
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
limx→a
f (x) = f (a).
Professor Christopher Hoffman Math 124
Evaluating Limits of Continuous Functions
Find the following limits.1 limx→3
√3πx+1x2−6 =
√9π+1
3
2 limy→−2sin√
3y+105y2−3 = sin(2)
17
3 limx→−2
√3cx+127cx+1 =
√12−6cx1−14c for any c < 2 and c 6= 1/14.
When evaluating limits in this course evaluating the function atthe point should always be your first approach.
If this fails then try pugging in numbers close to a. After youmake a guess you must justify your answer.
Professor Christopher Hoffman Math 124
Find
limh→0
(2 + h)3 − 8h
.
Plugging in h = 0 we get 00 .
limh→0
(2 + h)3 − 8h
= limh→0
(2 + h)3 − 8h
= limh→0
8 + 12h + 6h2 + h3 − 8h
= limh→0
12h + 6h2 + h3
h= lim
h→012 + 6h + h2
= 12.
If two functions agree everywhere except at one point a thenthe limits approaching a are the same.
Professor Christopher Hoffman Math 124
Findlimt→0
1/(7 + 3t)− 1/7t
.
Plugging in t = 0 we get 00 .
limt→0
1/(7 + 3t)− 1/7t
= limt→0
7− (7 + 3t)t(7 + 3t)(7)
= limt→0
−3tt(7 + 3t)7
= limt→0
−3(7 + 3t)7
= − 349.
Professor Christopher Hoffman Math 124
Rationalize the denominator
Findlim
t→16
16− t4−√
t.
Plugging in t = 16 we get 00 .
limt→16
16− t4−√
t= lim
t→16
(16− t)(4 +√
t)(4−
√t)(4 +
√t)
= limt→16
(16− t)(4 +√
t)(16− t)
= limt→16
4 +√
t
= 8
Professor Christopher Hoffman Math 124
Find
limx→0
(3x− 3
x3 + x
).
Plugging in x = 0 we get∞−∞.
limx→0
(3x− 3
x3 + x
)= lim
x→0
3(x2 + 1)− 3x3 + x
= limx→0
3x2
x3 + x
= limx→0
3xx2 + 1
=3 · 0
02 + 1= 0.
Professor Christopher Hoffman Math 124
Find
limx→3
x3 − 27x − 3
Plugging in x = 3 we get 00 .
limx→3
x3 − 27x − 3
= limx→0
(x − 3)(x2 + 3x + 9)x − 3
= limx→0
x2 + 3x + 9
= 27.
Professor Christopher Hoffman Math 124
Limit Laws
Suppose c is a constant and the limits
limx→a
f (x) and limx→a
g(x)
exist. Then1 limx→a
[f (x) + g(x)
]= limx→a f (x) + limx→a g(x)
2 limx→a[f (x)− g(x)
]= limx→a f (x)− limx→a g(x)
3 limx→a[cf (x)
]= c limx→a f (x)
4 limx→a[f (x) · g(x)
]= limx→a f (x) · limx→a g(x)
5
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
Find
limt→−1
(cos(
3πt + 1
)− 3)(3(t + 1)2
√−4t).
This looks like a mess. Let’s look at the graph.
Professor Christopher Hoffman Math 124
Find
limt→−1
(cos(
3πt + 1
)− 3)(3(t + 1)2
√−4t).
This looks like a mess. Let’s look at the graph.
Professor Christopher Hoffman Math 124
Here is the graph of the function
g(t) =(
cos(3π
t + 1)− 3
)(3(t + 1)2
√−4t)
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
limt→−1
−2(3(t + 1)2√−4t) = −2(3(−1 + 1)
√−4(−1) = 0
and
limt→−1
−4(3(t + 1)2√−4t) = −4(3(−1 + 1)
√−4(−1) = 0.
The function that we are interested in
g(t) =(
cos(3π
t + 1)− 3
)(3(t + 1)2
√−4t)
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.
TheoremIf f (x) ≤ g(x) ≤ h(x) in some interval around a and
limx→a
(f (x)) = limx→a
(h(x)) = L
thenlimx→a
(g(x)) = L
Professor Christopher Hoffman Math 124
Findlim
x→14
x − 14√14 + x
.
First we plug in x = 14 and get
14− 14√14 + 14
=0√28
= 0.
As this is a continuous function defined at 14 we have
limx→14
x − 14√14 + x
= 0.
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)x2 .
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 everypoint where it is defined.
It is defined everywhere the denominator is not zero, which iswhen x = 0 and x = −2.
It is important to keep signs right. We must be careful about theterms close to zero.
Professor Christopher Hoffman Math 124
As x → 0+, x2 approaches 0 from the right. Also when x → 0−,x2 approaches 0 from the right.
As x → 0 the numerator x − 1 approaches −1 and x + 2approaches 2.
Thus as x → 0 from either side the numerator is negative andthe denominator is small and positive. The fraction is thus largeand negative. Because of this
limx→0
x − 1(x + 2)x2 = −∞.
Professor Christopher Hoffman Math 124
As x → −2+ the numerator x − 1 approaches −3 and thedenominator approaches 0 from the positive side.
As the numerator is negative and the denominator is small andpositive the fraction is large and negative. Because of this
limx→−2+
x − 1(x + 2)x2 = −∞.
As x → −2− the numerator x − 1 approaches −3 and thedenominator approaches 0 from the negative side.
As the numerator is negative and the denominator is small andnegative the fraction is large and positive. Thus
limx→−2−
x − 1(x + 2)x2 =∞.
As the two one sided limits differ we have that
limx→−2
x − 1(x + 2)x2
does not exist.
Professor Christopher Hoffman Math 124
When we graph the function we can see that there are verticalasymptotes at x = −2 and x = 0.
Professor Christopher Hoffman Math 124
1 Find all vertical asymptotes of x−1(x+2)x2 .
2 Find the one sided limits of f (x) as x approaches theasymptotes.
The function is a rational function so it is continuous at everypoint where it is defined.It is defined everywhere the denominator is not zero, which iswhen x = 0 and x = −2.It is important to keep signs right. We must be careful about theterms close to zero.
Professor Christopher Hoffman Math 124
As x → 0+, x2 approaches 0 from the right. Also when x → 0−,x2 approaches 0 from the right.
As x → 0 the numerator x − 1 approaches −1 and x + 2approaches 2.
Thus as x → 0 from either side the numerator is negative andthe denominator is small and positive. The fraction is thus largeand negative. Because of this
limx→0
x − 1(x + 2)x2 = −∞.
Professor Christopher Hoffman Math 124
As x → −2+ the numerator x − 1 approaches −3 and thedenominator approaches 0 from the positive side.
As the numerator is negative and the denominator is small andpositive the fraction is large and negative. Because of this
limx→−2+
x − 1(x + 2)x2 = −∞.
As x → −2− the numerator x − 1 approaches −3 and thedenominator approaches 0 from the negative side.
As the numerator is negative and the denominator is small andnegative the fraction is large and positive. Thus
limx→−2−
x − 1(x + 2)x2 =∞.
As the two one sided limits differ we have that
limx→−2
x − 1(x + 2)x2
does not exist.
Professor Christopher Hoffman Math 124
Limit Laws
Suppose c is a constant and the limits
limx→a
f (x) and limx→a
g(x)
exist. Then1 limx→a
[f (x) + g(x)
]= limx→a f (x) + limx→a g(x)
2 limx→a[f (x)− g(x)
]= limx→a f (x)− limx→a g(x)
3 limx→a[cf (x)
]= c limx→a f (x)
4 limx→a[f (x) · g(x)
]= limx→a f (x) · limx→a g(x)
5 limx→af (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 writelim
x→∞f (x) = L
when the values of f (x) can be made arbitrarily close to L bytaking x sufficiently large. Similarly we write
limx→−∞
f (x) = L
when the values of f (x) can be made arbitrarily close to L bytaking 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) getscloser to the graph of y = 3 and we write
limx→∞
3− 2e−x = 3.
Professor Christopher Hoffman Math 124
Limits at∞ Graphically
As x gets more and more negative 3− 2e−x gets more andmore negative as well. We write
limx→−∞
3− 2e−x = −∞.
Professor Christopher Hoffman Math 124
Horizontal Asymptotes
The line y = L is a horizontal asymptote of the curve y = f (x) ifeither
limx→∞
f (x) = L or limx→−∞
f (x) = L.
Thus y = π/2 and y = −π/2 are horizontal asymptotes and
limx→−∞
tan−1(x) = π/2 and limx→∞
tan−1(x) = −π/2
Professor Christopher Hoffman Math 124
Limits of polynomials at∞
If r > 0 then limx→∞ x r =∞If r < 0 then limx→∞ x r = 0If 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
limx→∞
f (x) and limx→∞
g(x)
exist. Then
1 limx→∞[f (x) + g(x)
]= limx→∞ f (x) + limx→∞ g(x)
2 limx→∞[f (x)− g(x)
]= limx→∞ f (x)− limx→∞ g(x)
3 limx→∞[cf (x)
]= c limx→∞ f (x)
4 limx→∞[f (x) · g(x)
]= limx→∞ f (x) · limx→∞ g(x)
5 limx→∞f (x)g(x) =
limx→∞ f (x)limx→∞ g(x) if limx→∞ g(x) 6= 0
Professor Christopher Hoffman Math 124
Extended Real Numbers
When working with limits it is useful to use the extended realnumbers.
∞ ·∞ =∞∞+∞ =∞∞+ a =∞a∞ =∞ if a > 0a∞ = −∞ if a < 0∞−∞ =???
Professor Christopher Hoffman Math 124
Rational Functions
Find the following limits.limx→∞(−5x3 + 2(r − 1)x)
limx→∞x3/2+2πx7−2x
1
limx→−∞−5x3+2(r−1)x√
x4+1
Professor Christopher Hoffman Math 124
Polynomials
With polynomials the largest term dominates. To see this wefactor out the largest power of x .
limx→∞
(−5x3 + 2(r − 1)x) = limx→−∞
x3(−5 + 2(r − 1)/x2)
= limx→−∞
x3 limx→−∞
(−5 + 2(r − 1)/x2)
= (−∞)(−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 largestpower of the denominator.
limx→∞
x3/2 + 2πx7 − 2xx2 − 1
= limx→∞
(1/x2)(x3/2 + 2πx7 − 2x)(1/x2)(x2 − 1)
= limx→∞
x−1/2 + 2πx5 − 2/x)1− 1/x2
=limx→∞(x−1/2 + 2πx5 − 2/x)
limx→∞(1− 1/x2)
=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 x2 =√
x4.
limx→−∞
−5x3 + 2(r − 1)x√x4 + 1
= limx→−∞
(1/x2)(−5x3 + 2(r − 1)x)(1/x2)
√x4 + 1
= limx→−∞
−5x + 2(r − 1)/x√1 + 1/x4
=limx→−∞(−5x + 2(r − 1)/x)
limx→−∞√
1 + 1/x4
=−5(−∞) + 0
1= −∞
Professor Christopher Hoffman Math 124
limx→∞
(√x2 + 3x + 2− x
)We rationalize the numerator by multiplying the expression by
√x2 + 3x + 2 + x√x2 + 3x + 2 + x
limx→∞
(√x2 + 3x + 2− x
)= lim
x→∞
(√x2 + 3x + 2− x
) √x2 + 3x + 2 + x√x2 + 3x + 2 + x
= limx→∞
x2 + 3x + 2− x2√
x2 + 3x + 2 + x
= limx→∞
3x + 2√x2 + 3x + 2 + x
Professor Christopher Hoffman Math 124
Then we multiply numerator and denominator by 1/x .
limx→∞
(√x2 + 3x + 2− x
)= lim
x→∞
(3x + 2)(1/x)(√
x2 + 3x + 2 + x)(1/x)
= limx→∞
3 + 2/x(√
1 + 3/x + 2/x2 + 1)
=limx→∞(3 + 2/x)
limx→∞(√
1 + 3/x + 2/x2 + 1)
=3
1 + 1
=32
Professor Christopher Hoffman Math 124
For all real numbers a > 0,b and c find
limx→∞
(√ax2 + bx + c − 5x
)and lim
x→−∞
(√ax2 + bx + c − 5x
)As in the previous problem we will multiply the expression times
√ax2 + bx + c + 5x√ax2 + bx + c + 5x
and then factor out the largest power of the denominator.
Professor Christopher Hoffman Math 124
limx→∞
(√ax2 + bx + c − 5x
)= lim
x→∞(√
ax2 + bx + c − 5x)√
ax2 + bx + c + 5x√ax2 + bx + c + 5x
= limx→∞
ax2 + bx + c − 25x2√
ax2 + bx + c + 5x
= limx→∞
(a− 25)x2 + bx + c√ax2 + bx + c + 5x
= limx→∞
(1/x)(a− 25)x2 + bx + c)(1/x)(
√ax2 + bx + c + 5x)
= limx→∞
(a− 25)x + b + c/x√a + b/x + c/x2 + 5
=limx→∞((a− 25)x + b + c/x)
limx→∞(√
a + b/x + c/x2 + 5)
=limx→∞((a− 25)x + b + 0)√
aNow we have to break the answer up into cases.
Professor Christopher Hoffman Math 124
limx→∞
(√ax2 + bx + c − 5x
)=
limx→∞((a− 25)x + b + 0)√a
If a > 25 then the leading term has a positive coefficient andthe limit approaches∞.
If a < 25 then the leading term has a negative coefficient andthe 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
Findlim
x→∞
(√4x4 + 5x3 + 10− (2x2 + x)
).
Again we will multiply by the conjugate and divide by thehighest power of x in the denominator.
limx→∞
(√4x4 + 5x3 + 10− (2x2 + x)
)= lim
x→∞
(√4x4 + 5x3 + 10− (2x2 + x)
) √4x4 + 5x3 + 10 + (2x2 + x)√4x4 + 5x3 + 10 + (2x2 + x)
= limx→∞
(4x4 + 5x3 + 10)− (2x2 + x)2√
4x4 + 5x3 + 10 + (2x2 + x)
= limx→∞
(4x4 + 5x3 + 10)− (4x4 + 4x3 + x2)√4x4 + 5x3 + 10 + (2x2 + x)
= limx→∞
x3 − x2 + 10√4x4 + 5x3 + 10 + (2x2 + x)
Professor Christopher Hoffman Math 124
= limx→∞
x − 1 + 10/x2
2
√4 + 5/x + 10/x4 + 2 + 1/x2
= ∞
Professor Christopher Hoffman Math 124
1 Find all vertical asymptotes of x−1(x+2)x2 .
2 Find all horizontal asymptotes of x−1(x+2)x2 .
3 Find the one sided limits of f (x) as x approaches theasymptotes.
The function is a rational function so it is continuous at everypoint where it is defined.It is defined everywhere the denominator is not zero, which iswhen x = 0 and x = −2.It is important to keep signs right. We must be careful about theterms close to zero.
Professor Christopher Hoffman Math 124
As x → 0+, x2 approaches 0 from the right. Also when x → 0−,x2 approaches 0 from the right.
As x → 0 the numerator x − 1 approaches −1 and x + 2approaches 2.
Thus as x → 0 from either side the numerator is negative andthe denominator is small and positive. The fraction is thus largeand negative. Because of this
limx→0
x − 1(x + 2)x2 = −∞.
Professor Christopher Hoffman Math 124
As x → −2+ the numerator x − 1 approaches −3 and thedenominator approaches 0 from the positive side.
As the numerator is negative and the denominator is small andpositive the fraction is large and negative. Because of this
limx→−2+
x − 1(x + 2)x2 = −∞.
As x → −2− the numerator x − 1 approaches −3 and thedenominator approaches 0 from the negative side.
As the numerator is negative and the denominator is small andnegative the fraction is large and positive. Thus
limx→−2−
x − 1(x + 2)x2 =∞.
As the two one sided limits differ we have that
limx→−2
x − 1(x + 2)x2
does not exist.
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 = 42 has vertical asymptotes x = −3 and x = 43 limx→4 f (x) =∞4 limx→−3 f (x) does not exist.
Professor Christopher Hoffman Math 124
Graphical limitsComposition of functions.Change of variables.
Professor Christopher Hoffman Math 124