CalcI Infinite Limits Solutions

8/6/2019 CalcI Infinite Limits Solutions

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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 havemore or less detail than other solutions. The level of detail in each solution will depend up onseveral issues. If the section is a review section, this mostly applies to problems in the firstchapter, there will probably not be as much detail to the solutions given that the problems reallyshould 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 youve reached the level of working the harder problems then you will probably already understand the basics fairly well and wont need all theexplanation.

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 theweb pages that havent been removed here. These issues may make the solutions a little difficultto follow at times, but they should still be readable.

Infinite Limits

1. For ( ) ( )59

3 f x

x=

-evaluate the indicated limits, if they exist.

(a) ( )3

lim x

f x-

(b) ( )3

lim x

f x+

(c) ( )3

lim x

f x

(a) ( )3

lim x

f x-

Lets start off by acknowledging that for 3 x - 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,3 0 x -

but we can actually go a little farther. Because we know that 3 x < we also know that,3 0 x - <
http://tutorial.math.lamar.edu/terms.aspxhttp://tutorial.math.lamar.edu/terms.aspx


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Calculus I


More compactly, we can say that in the limit we will have,

3 0 x --

Raising this to the fifth power will not change this behavior and so, in the limit, the denominator will be,

( )53 0 x --

We can now do the limit of the function. In the limit, the numerator is a fixed positive constantand the denominator is an increasingly small negative number. In the limit, the quotient mustthen be an increasing large negative number or,

( )539

lim3 x x

-

= -

-

Note that this also means that there is a vertical asymptote at 3 x = .

(b) ( )3

lim x

f x+

Lets start off by acknowledging that for 3 x + 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,3 0 x -

and because we know that 3 x > we also know that,

3 0 x- >


3 0 x +-

Raising this to the fifth power will not change this behavior and so, in the limit, the denominator will be,

( )53 0 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. In the limit, the quotient must then be an increasing large positive number or,

( )539

lim3 x x

+

=

-

Note that this also means that there is a vertical asymptote at 3 x = , which we already knew fromthe first part.


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Calculus I


(c) ( )3

lim x

f x

In this case we can see from the first two parts that,

( ) ( )3 3

lim lim x x

f x f x- +

and so, from our basic limit properties we can see that ( )3

lim x

f x

does not exist .

For the sake of completeness and to verify the answers for this problem here is a quick sketch of the function.

2. For ( )2

6t

h t t =

+ evaluate the indicated limits, if they exist.

(a) ( )6

limt

h t -

-

(b) ( )6

limt

h t +

-

(c) ( )6

limt

h t -

(a) ( )6

limt

h t -

-

Lets start off by acknowledging that for 6t -

- we know 6t < - .

For the numerator we can see that, in the limit, we will get -12.

The denominator takes a little more work. Clearly, in the limit, we have,6 0t +

but we can actually go a little farther. Because we know that 6t < - we also know that,6 0t + <



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Calculus I


6 0t -+

So, in the limit, the numerator is approaching a negative number and the denominator is anincreasingly small negative number. The quotient must then be an increasing large positivenumber or,

6

2lim6t

t t --

= +

Note that this also means that there is a vertical asymptote at 6t = - .

(b) ( )6

limt

h t +

-

Lets start off by acknowledging that for 6t + - we know 6t > - .

For the numerator we can see that, in the limit, we will get -12.

The denominator will also work similarly to the first part. In the limit, we have,6 0t +

but we can actually go a little farther. Because we know that 6t > - we also know that,6 0t + >


6 0t ++

So, in the limit, the numerator is approaching a negative number and the denominator is an

increasingly small positive number. The quotient must then be an increasing large negativenumber or,

6

2lim

6t t t + -

= -+

Note that this also means that there is a vertical asymptote at 6t = - , which we already knewfrom the first part.

(c) ( )6

limt

h t -

In this case we can see from the first two parts that,( ) ( )

6 6lim lim

t t h t h t

- +- -


limt

h t -

does not exist .



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Calculus I


3. For ( )( )2

3

1

z g z

z

+=

+evaluate the indicated limits, if they exist.

(a) ( )1

lim z

g z -

-

(b) ( )1

lim z

g z +

-

(c) ( )1

lim z

g z -

(a) ( )1

lim z

g z -

-

Lets start off by acknowledging that for 1 z -

- we know 1 z < - .

For the numerator we can see that, in the limit, we will get 2.

Now lets take care of the denominator. In the limit, we will have,1 0 z

-+

and upon squaring the 1 z + we see that, in the limit, we will have,

( )21 0 z ++

So, in the limit, the numerator is approaching a positive number and the denominator is anincreasingly small positive number. The quotient must then be an increasing large positivenumber or,

( )213

lim1 z

z

z -

-

+=

+

Note that this also means that there is a vertical asymptote at 1 z = - .

(b) ( )1

lim z

g z +

-


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Calculus I


Lets start off by acknowledging that for 1 z +

- we know 1 z > .


Now lets take care of the denominator. In the limit, we will have,1 0 z ++

and upon squaring the 1 z + we see that, in the limit, we will have,

( )21 0 z ++


( )213

lim1 z

z

z +

-

+=

+

Note that this also means that there is a vertical asymptote at 1 z = - , which we already knewfrom the first part.

(c) ( )1

lim z

g z -


( ) ( )1 1

lim lim z z

g z g z - +

- -

= =

and so, from our basic limit properties we can see that,

( )1lim z

g z -

=



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Calculus I


4. For ( ) 274

x g x

+=

-evaluate the indicated limits, if they exist.

(a) ( )2

lim x

x-

(b) ( )2

lim x

x+

(c) ( )2

lim x

x

(a) ( )2

lim x

x-

Lets start off by acknowledging that for 2 x - we know 2 x < .


Now lets take care of the denominator. First, we know that if we square a number less than 2(and greater than -2, which it is safe to assume we have here because were doing the limit) wewill get a number that is less that 4 and so, in the limit, we will have,

2 4 0 x --

So, in the limit, the numerator is approaching a positive number and the denominator is anincreasingly small negative number. The quotient must then be an increasing large negativenumber or,

22

7lim

4 x x x-

+= -

-

Note that this also means that there is a vertical asymptote at 2 x = .

(b) ( )2

lim x

x+

Lets start off by acknowledging that for 2 x + we know 2 x > .


Now lets take care of the denominator. First, we know that if we square a number greater than 2we will get a number that is greater than 4 and so, in the limit, we will have,

2 4 0 x +-


22

7lim

4 x x x+

+=

-


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Calculus I


Note that this also means that there is a vertical asymptote at 2 x = , which we already knew fromthe first part.

(c) ( )2

lim x

g x


( ) ( )2 2

lim lim x x

g x g x- +


lim x

g x

does not exist .


As were sure that you had already noticed there would be another vertical asymptote at 2 x = - for this function. For the practice you might want to make sure that you can also do the limits for that point.

5. For ( ) ( )lnh x x= - evaluate the indicated limits, if they exist.

(a) ( )0

lim x

h x-

(b) ( )0

lim x

h x+

(c) ( )0

lim x

h x

Hint : Do not get excited about the x- inside the logarithm. Just recall what you know aboutnatural logarithms, where they exist and dont exist and the limits of the natural logarithm at

0 x = .

(a) ( )0

lim x

h x-


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Okay, lets start off by acknowledging that for 0 x - we know 0 x < and so 0 x- > or,0 x +-

What this means for us is that this limit does make sense! We know that we cant have negativearguments in a logarithm, but because of the negative sign in this particular logarithm that meansthat we can use negative xs in this function (positive xs on the other hand will now cause

problems of course).

By Example 6 in the notes for this section we know that as the argument of a logarithmapproaches zero from the right (as ours does in this limit) then the logarithm will approach - .

Therefore, the answer for this part is,

( )0

lim ln x

x-

= -

(b) ( )0lim x h x+ In this part we know that for 0 x + we have 0 x > and so 0 x- < . At this point we can stop.We know that we cant have negative arguments in a logarithm and for this limit that is exactly

what well get and so ( )0

lim x

h x+

does not exist .

(c) ( )0

lim x

h x

The answer for this part is ( )0

lim x

h x

does not exist . We can use two lines of reasoning to justify

this.

First, we are unable to look at both sides of the point in question and so there is no possible wayfor the limit to exist.

The second line of reasoning is really the same as the first, but put in different terms. From thefirst two parts that,

( ) ( )0 0

lim lim x x

h x h x- +


lim x

h x

does not exist .



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6. For ( ) ( )tan R y y= evaluate the indicated limits, if they exist.

(a) ( )32

lim y

R yp

-

(b) ( )32

lim y

R yp

+

(c) ( )32

lim y

R yp

Hint : Dont forget the graph of the tangent function.

(a) ( )32

lim y

R yp

-

The easiest way to do this problem is from the graph of the tangent function so here is a quick

sketch of the tangent function over several periods.

From the sketch we can see that,

( )32

lim tan y

yp

-

=


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(b) ( )32

lim y

R yp

+

From the graph in the first part we can see that,

( )32

lim tan y

yp

+

= -

(c) ( )32

lim y

R yp

From the first two parts that,

( ) ( )3 32 2

lim lim y y

R y R yp p

- +


lim y

R yp

does not exist .

7. Find all the vertical asymptotes of ( )( )4

7

10 3

x f x =

-.

Hint : Remember how vertical asymptotes are defined and use the examples above to helpdetermine where they are liable to be for the given function. Once you have the locations for the

possible vertical asymptotes verify that they are in fact vertical asymptotes.

SolutionRecall that vertical asymptotes will occur at a= if any of the limits (one-sided or overall limit)at a= are plus or minus infinity.

From previous examples we can see that for rational expressions vertical asymptotes will occur where there is division by zero. Therefore it looks like the only possible vertical asymptote will

be at 103 x = .

Now lets verify that this is in fact a vertical asymptote by evaluating the two one-sided limits,

( ) ( )10 103 34 47 7

lim and lim10 3 10 3 x x

x x

x- +

- -

In either case as 103 x (from both left and right) the numerator goes to703 .

For the one-sided limits we have the following information,10 10 103 3 3

10 10 103 3 3

0 10 3 0

0 10 3 0

x x x x

x x x x

-

+

< - > - >

> - < -


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Also, note that for xs close enough to -5 (which because were looking at 5 x - is safeenough to assume), we will have 9 0 x - < .

So, in the left-hand limit, the numerator is a fixed negative number and the denominator is positive (a product of two negative numbers) and increasingly small. Likewise, for the right-hand

limit, the denominator is negative (a product of a positive and negative number) and increasinglysmall. Therefore, we will have,

( )( ) ( )( )5 58 8

lim and lim5 9 5 9 x x x x x x- + - -

- -= - =

+ - + -

Now for 9 x = . Again, the numerator is a constant -8. We also have,

9 9 9 0

9 9 9 0

x x x

x x x

-

+

< - - >

Finally, for xs close enough to 9 (which because were looking at 9 x is safe enough toassume), we will have 5 0 x + > .

So, in the left-hand limit, the numerator is a fixed negative number and the denominator isnegative (a product of a positive and negative number) and increasingly small. Likewise, for theright-hand limit, the denominator is negative (a product of two negative numbers) andincreasingly small. Therefore, we will have,

( )( ) ( )( )9 98 8

lim and lim5 9 5 9 x x x x x x- +

- -= = -

+ - + -

So, as all of these limits show we do in fact have vertical asymptotes at 5 x = - and 9 x = .

CalcI Infinite Limits Solutions

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