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Calculus I
© 2007 Paul Dawkins
1
http://tutorial.math.lamar.edu/terms.aspx
Preface
Here are the solutions to the practice problems for my Calculus I notes. Some solutions will have
more or less detail than other solutions. The level of detail in each solution will depend up on
several issues.
If the section is a review section, this mostly applies to problems in the first
chapter, there will probably not be as much detail to the solutions given that the problems really
should be review.
As the difficulty level of the problems increases less detail will go into the
basics of the solution under the assumption that if you’ve reached the level of working the harder
problems then you will probably already understand the basics fairly well and won’t need all the
explanation.
This document was written with presentation on the web in mind.
On the web most solutions are
broken down into steps and many of the steps have hints.
Each hint on the web is given as a
popup however in this document they are listed prior to each step.
Also, on the web each step can
be viewed individually by clicking on links while in this document they are all showing.
Also,
there are liable to be some formatting parts in this document intended for help in generating the
web pages that haven’t been removed here. These issues may make the solutions a little difficult
to follow at times, but they should still be readable.
Infinite
Limits
1. For
()
5
9
3
fx
x
=

evaluate the indicated limits, if they exist.
(a)
(
)
3
lim
x

ﬁ
(b)
(
)
3
lim
x
+
ﬁ
(c)
(
)
3
lim
x
ﬁ
(a)
(
)
3
lim
x

ﬁ
Let’s start off by acknowledging that for
3
x

ﬁ
we know
3
x
<
.
For the numerator we can see that, in the limit, it will just be 9.
The denominator takes a little more work.
Clearly, in the limit, we have,
30
x
ﬁ
but we can actually go a little farther.
Because we know that
3
x
<
we also know that,
x
<
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View Full Document Calculus I
© 2007 Paul Dawkins
2
http://tutorial.math.lamar.edu/terms.aspx
More compactly, we can say that in the limit we will have,
30
x

ﬁ
Raising this to the fifth power will not change this behavior and so, in the limit, the denominator
will be,
()
5
x

We can now do the limit of the function.
In the limit, the numerator is a fixed positive constant
and the denominator is an increasingly small negative number.
In the limit, the quotient must
then be an increasing large negative number or,
5
3
9
lim
3
x
x

ﬁ
= ¥

Note that this also means that there is a vertical asymptote at
3
x
=
.
(b)
( )
3
lim
x
fx
+
ﬁ
Let’s start off by acknowledging that for
3
x
+
ﬁ
we know
3
x
>
.
As in the first part the numerator, in the limit, it will just be 9.
The denominator will also work similarly to the first part. In the limit, we have,
x
and because we know that
3
x
>
we also know that,
x
>
More compactly, we can say that in the limit we will have,
x
+
Raising this to the fifth power will not change this behavior and so, in the limit, the denominator
will be,
5
x
+
We can now do the limit of the function.
In the limit, the numerator is a fixed positive constant
and the denominator is an increasingly small positive number.
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This note was uploaded on 02/05/2011 for the course CALCULUS/C 202 taught by Professor Tadius during the Spring '11 term at Benedictine KS.
 Spring '11
 Tadius

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