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The 5th Wave_By RichTennant

‘VJV\at exactkj are vre saying here?’

In this part…

/saac Newton and Gottfried Leibniz, co-inventors of calculus in the late 17th century, did calculus without the solid foundation provided by limits and continuity. Now, since Newton and Leibniz were merely two of the greatest geniuses who ever lived, you don’t want to do math at their lowly level, do you? Of course not! You want your calculus to be mathematically rigorous, and for that you need to master limits and continuity — despite their almost complete lack of practical importance. Now, don’t blame me for this — heck, if practical relevance is your standard, you might as well drop out of school all together. Actually, the study of limits (and to a much lesser extent continuity) can

Be fascinating. And in the event that you hail from the show-me state, limits will allow you to prove that in a race against a tortoise where you give the tortoise a head start,

You actually will catch up to and then pass the tortoise. The things that mathematics allows you to accomplish!

Chapter 3

Ou can use ordinary algebra and geometry when the things in a math problem Aren’t

Changing (sort of) and when lines are Straight. But you need calculus when things Are Changing (these changing things are often represented as Curves). For example, you need calculus to analyze something like the motion of the space shuttle during the beginning of its flight because its acceleration is changing every split second.

Ordinary algebra and geometry fall short for such things because the algebra or geometry formula that works one moment no longer works a millionth of a second later. Calculus, on the other hand, chops up these constantly changing things — like the motion of the space shuttle — into such tiny bits (actually infinitely small bits) that within each bit, things don’t change. Then you Can Use ordinary algebra and geometry.

Limits Are the "magical" trick or tool that does this chopping up of something into infinitely small bits. It’s the mathematics of limits that makes calculus work. Limits are so essential

That the formal definitions of the derivative and the definite integral both involve limits.

If — when your parents or other adults asked you, "What do you want to be when you grow up?" — you responded, "Why, a mathematician, of course," then you may ultimately spend a great deal of time thoroughly studying the deep and rich subtleties of Continuity. For the rest of you, the concept of continuity is a total no-brainer. If you can draw a graph without lifting your pen or pencil from the page, the graph is Continuous. If you can’t — because there’s a break in the graph — then the graph is not continuous. That’s all there is to it. By the way, there are some subtle and technical connections between limits and continuity (which I don’t want to get into), and that’s why they’re in the same chapter. But, be honest now, did you buy this book because you were dying to learn about mathematical subtleties and

Technicalities?

Digesting the Definitions: Limit and Continuity

In This Chapter

► The mathematical mumbo jumbo of limits and continuity

► When limits exist and Don’t Exist

► Discontinuity. .. or graphus interruptus

This short section covers a couple formal definitions and a couple other things you need to

Know about limits and continuity. Here’s the formal, three-part definition of a limit:

For a function f (X) And a real number a, lim f (X) Exists if and only if

1. lim f (x) exists. In other words, there must be a limit from the left.

2. lim f (x) exists. There must be a limit from the right.

3. lim f (x) = Liim f (x) The limit from the left must equal the limit from the right. (Note that this definition does not apply to limits as x approaches infinity or negative

And here’s the definition of continuity: A function f (x) is continuous at a point x = A If three conditions are satisfied:

1. f (a) Is defined.

2. limf (x) exists.

3. f(a) = limf(x).

Using these definitions and Figure 3-1, answer problems 1 through 4.

Infinity.)

15

Interruptus: A function with many discontinuities.

Figure 3-1:

Graphus

F

X

1. At which of the following x-values are all

Three requirements for the existence of a limit satisfied, and what is the limit at

Those x-values? x = -2, 0, 2, 4, 5, 6, 8, 10,

And11.

Solve It

2. For the x-values at which all three limit

Requirements are not met, state which of the three requirements are not satisfied. Ifone or both one-sided limits exist at any of these – values, give the value of the one-sided limit.

Solve It

3. At which of the x-values are all three requirements for continuity satisfied?

Solve It

Ft. For the rest of the x-values, state which of the three continuity requirements are not

Satisfied.

Solve It

Taking a Closer Look: Limit and Continuity Graphs

In this section, you get more practice at solving limit and continuity problems visually. Then in Chapter 4, you solve limit problems numerically (with your calculator) and symbolically (with algebra).

Use Figure 3-2 to answer problems 5 through 10.

Q. Lim f (x) = ?

A. Lim ° (X) = 2 Because f (0) = 2 and because

Fis continuous there, the limit must equal

The function value. Whenever a function

Passes through a point and there’s no discontinuity at the point, the limit equals the function value.

Q. Limn f (x) = ?

A. Lim ° (X) = 2 Because there’s a hole at

(13, 2). The limit at a hole is the height ofthe hole.

5. Limf (x) = ?

Solve It

6. Lim f (x) = ?

Solve It

7. Lim f (x) = ?

Solve It

8. Lim f (x) = ?

1 Q V ‘

Solve It

9. Lim ° (x) = ?

Solve It

10. List the x-coordinates of all points of discontinuity of ° and state the type of discontinuity — removable, jump, or infinite.

Solve It

11. Lim sinx = ? See the following graph.

12. Lim= ? See the following graph of y = J1.

Solve /t

Y

Y

X

X

Solve /t

Solutions for Limits and Continuity

MM At which of the following x-values are all three requirements for the existence of a limit satisfied, and what is the limit at those x-values? x = -2, 0, 2, 4, 5, 6, 8, 10, and 11.

At 0, the limit is 2.

At 4, the limit is 5.

At 8, the limit is 3.

At 10, the limit is 5.

" Jill

To make a long story short, a limit exists at a particular x-value of a curve when the curve is Heading toward An exact y-value and keeps Heading toward That y-value as you continue to zoom

In on the curve at the X-value. The curve must head toward that Y-value from the right and from

The left (unless the limit is one where x approaches infinity). I emphasize "heading toward" because what happens precisely at the given x-value isn’t relevant to this limit inquiry. That’s why there is a limit at a hole like the ones at x = 8 and 10.

MM For the rest of the x-values, state which of the three limit requirements are not satisfied. If one or both one-sided limits exist at any of these x-values, give the value of the one-sided limit.

At -2 and 5, all three conditions fail.

At 2, 6, and 11, only the third requirement is not satisfied. At 2, the limit from the left equals 5 and the limit from the right equals 3. At 6, the limit from the left is 2 and the limit from the right is 3. Finally, lim ° (x) = 3 and lim ° (X) = 5.

X-11" X-11′

D At which of the x-values are all three requirements for continuity satisfied?

The function in Figure 3-1 is continuous at 0 and 4. The common-sense way of thinking about

Continuity is that a curve is continuous wherever you can draw the curve without taking your

Pen off the paper. It should be obvious that that’s true at 0 and 4, but not at any of the other

Listed X-values.

MM For the rest of the x-values, state which of the three continuity requirements are not satisfied. All listed x-values other than 0 and 4 are points of discontinuity. A discontinuity Is just a high -

Falutin’ calculus way of saying a gap. If you’d have to take your pen off the paper at some point

When drawing a curve, then the curve has a discontinuity there.

At 5 and 11, all three conditions fail.

At -2, 2, and 6, continuity requirements 2 and 3 are not satisfied.

At 10, requirements 1 and 3 are not satisfied. At 8, requirement 3 is not satisfied.

D Lim ° (x) does not exist (DNE) because there’s a vertical asymptote at -7. Or, because F (x) Approaches - Oo From the left and from the right, you could say the limit equals - Oo.

MM Lim ° (x) does not exist because the limit from the left does not equal the limit from the

X — 5

Right. Or you could say that the limit DNE because there’s a jump discontinuity at X = 5.

D Lini ° (x) = 5 because, like the second example problem, the limit at a hole is the height of the hole. The fact that F (18) is undefined is irrelevant to this limit question.

D Lim °(x) = 4 because ° (5) = 4 and °is continuous from the left at (5, 4).

D Lim ° (x) = 6. This question is just like problem 8 except that there’s a hollow dot — instead

Of a solid one — at (5, 6). But the hollow dot at (5, 6) is irrelevant to the limit question — just like with question 7 where the hole was irrelevant.

E3 List the x-coordinates of all points of discontinuity of F And state whether the points of discontinuity are removable or non-removable, and state the type of discontinuity — removable, jump,

Or infinite.

At x = -7, the vertical asymptote, there is a non-removable, infinite discontinuity. At x = 5 there’s a non-removable, jump discontinuity.

At x = 13 and x = 18 there are holes which are removable discontinuities. Though infinitely

Small, these are nevertheless discontinuities. They’re "removable" discontinuities because you

Can "fix" the function by plugging the holes. ID Lim sinx does not exist. There’s no limit as X Approaches infinity because the curve oscillates

It never settles down to one exact y-value. (The three-part definition of a limit does not apply to

Limits at infinity.)

H Lim x = 0. In contrast to sinx, this function does hone in on a single value; as you go out

X — Oo

Further and further to the right, the function gets closer and closer to zero, so that’s the limit.

In This Chapter

► Algebra, schmalgebra

► Calculators — taking the easy way out Making limit sandwiches

► Infinity — "Are we there yet?"

► Conjugate multiplication — sounds R Rated, but it’s strictly PG

/n this chapter, you practice two very different methods for solving limit problems: using algebra and using your calculator. Learning the algebraic techniques are valuable for two reasons. The first, Incredibly Important reason is that the mathematics involved in the algebraic methods is beautiful, pure, and rigorous; and, second — something so trivial that perhaps I shouldn’t mention it — you’ll be tested on it. Do I have my priorities straight or what? The calculator techniques are useful for several reasons: 1) You can solve some limit problems on your calculator that are either impossible or just very difficult to do with algebra,

2) You can check your algebraic answers with your calculator, and 3) Limit problems can be solved with a calculator when you’re not required to show your work — like maybe on a

Multiple choice test.

But before we get to these two major techniques, how about a little rote learning. A few limits are a bit tricky to justify or prove, so to make life easier, simply commit them to memory. Here they are:

Limc = c

(y = c is a horizontal

Line, so the limit equals C Regardless of the

Arrow-number

The

Constant after the arrow.)

\S Lim \ = Oo LimF = — oo

\S Lim F = 0

\S Lim X = O

X — — oo

\S Limsinx = 1

X — 0

X

Cos —

X

]/* lim

\S Lim [ 1 +

1

0

—

Solving Limits ©ith Algebra

^BE/f You can solve limit problems with several algebraic techniques. But your first step should always be plugging the arrow-number into the limit expression. If you get a number, that's the answer. You're done. You're also done if plugging in the arrow-number gives you

\S A number or infinity or negative infinity over zero, like 0, or ^; in these cases the limit does not exist (DNE). 0 0

Zero over infinity; the answer is zero.

When plugging in fails because it gives you 0, you've got a real limit problem, and you have to convert the fraction into some expression where plugging in Does Work. Here

Are some algebraic methods you can try:

\S FOILing \S Factoring

Finding the least common denominator Canceling

\S Simplification

Conjugate multiplication The following examples each use a different method to solve the limit.

Q. Evaluate lim

16 — X 4 — Jx

A. The limit is 8.

1. Try plugging 16 into x

No good.

2. Multiply numerator and denominator by the conjugate of 4 — JX, namely 4 + /X.

The conjugate of a two-term expression has a plus sign instead of a minus sign — or vice-versa.

Lim

1 c

(16 — X)

(4 + Jx)

3. FOIL the conjugates and simplify.

(16 — X)(4 +fx) Because,

=lim

V _ lfi

42 — r'\ Of course, 4 I/X J (a — b)(a + b)■■

(16 — X)(4 +fx)

A2—b2

= lim - , ,

X —16 (16 — X)

= lim (4 + Jx)

4. Now you can cancel and then

Plug in.

4+16

Note that while plugging in did not work in Step 1, it did work in the final step. That's your goal: to change the original

Expression — usually by canceling — so

That plugging in works.

Q. A.

What's lim2

——2 2+ —2

The limit is 33.

1. Try plugging -2 into x — that gives you ^, So on to plan B.

2. Factor and cancel.

= (X + 2)( X — 3) X1—mm2 (X + 2)(X — 1)

= lim 7-rf

3. Cancel now and plug in.

—2—3

1. Lim

— 3

Solve It

2. Lim

—1

—1 2+ —2

Solve It

3. Lim

X + 2 2 X3 + 8

T. L im2

—2 4x2 + 5x — 6

Solve It

Solve It

9

X

5. Lim

X — 9 3 — J~x

Solve It

6. Lim

V _ in

Jx—5—y5

X—T0

Solve It

7. Lim

—0

Solve It

Cos —

8. Lim

1 — 1 .1-—I-

2 X — 2

Solve /t

1

9.

Lim

X — 0 1

6 X-

Solve /t

10. Lim

Solve /t

Sin

Lim

X

0 sin3

Solve /t

*12. Lim

Solve /t

Tan

X

6

X

Pulling Out \lour Calculator: UsefuL "Cheating"

Your calculator is a great tool for understanding limits. It can often give you a better feel for how a limit works than the algebraic techniques can. A limit problem asks you to determine what the y-value of a function is zeroing in on as the x-value approaches a particular number. With your calculator, you can actually witness the process and the

Result. You can solve a limit problem with you calculator in three different ways.

Method I. First, store a number into X That's extremely close to the arrow-number, enter the limit expression in the home screen, and hit Enter. If you get a result really close to a round number, that's your answer — you're done. If you have any doubt about the answer, just store another number into X That's even closer to the arrow-number and hit Enter Again. This will likely give you a result even closer to the same round number — that's it, you've got it. This method can be the quickest, but it often doesn't give you a good feel for how the y-values zero in on the result. To get a better picture of this process, you can store three or four numbers into X (one after another),

Each a bit closer to the arrow-number, and look at the sequence of results.

Method II. Enter the limit expression in graphing or "y =" mode, go to Table Setup, Set Tblstart To the arrow-number, and set ATbl To something small like 0.01 or 0.001. When you look at the table, you'll often see the y-values getting closer and closer to the limit

Answer as hones in on the arrow-number. If it's not clear what the Y-values are

Approaching, try a smaller increment for the ATbl Number. This method often gives you a good feel for what's happening in a limit problem.

Method III. This method gives you the best Visual Understanding of how a limit works. Enter the limit expression in graphing or "y =" mode. (If you're using the second method, you may want to try this third method at the same time.) Next, graph the function, and then go into the Window And tweak the Xmin, xmax, ymin, And Ymax Settings, if necessary, so that the part of the function corresponding to the arrow-number

Is within the viewing window. Use the Trace Feature to trace along the function until you get Close To the arrow-number. You can't trace exactly Onto The arrow-number because

There's a little hole in the function there, the height of which, by the way, is your answer. When you trace close to the arrow-number, the Y-value will get close to the

Limit answer. Use the ZoomBox Feature to draw a little box around the part of the graph

Containing the arrow-number and zoom in until you see that the Y-values are getting very close to a round number — that's your answer.

Q. A.

Evaluate lim

5 —6

—6sin( —6)

The answer is 7.

Method I.

1. UsetheSTO into X.

Button to store 6.01

2. Enter

X2 — 5x — 6

On the home

Sin (x — 6) screen and hit Enter. (Note: You must be in Radian Mode.)

This gives you a result of -7.01, suggesting that the answer is 7.

3. Repeat Steps 1 and 2 with 6.001

Stored into X.

This gives you a result of -7.001.

4. Repeat Steps 1 and 2 with 6.0001 stored into X.

This gives you a result of -7.0001.

Because the results are obviously honing in on the round number of 7,

That's your answer.

Method II. 1. Enter

5x — 6 •

Sin (x — 6) "§ =" mode.

In graphing or

2. Go to Table Setup And set TblStart To the arrow-number, 6, and ATbl To 0.01.

3. Go to the Table and you'll see the §-values getting closer and closer to 7 as you scroll toward x = 6 from

Above and below 6.

So 7 is your answer.

Method III. 1. Enter

X2 — 5x — 6

Sin(x

Mode again.

6)

In graphing

2. Graph the function. For your first

Viewing, ZoomStd, ZoomFit, And ZoomTrig (for expressions containing trig functions) are good windows to try.

For this funny function, none of these three window options works

Very well, but ZoomStd Is the best.

3. Trace close to X = 6 and you'll see

That § is near 7. Use ZoomBox To

Draw a little box around the point

(6, 7) then hit Enter.

4. Trace near X = 6 on this zoomed-in

Graph until you get very near to

X = 6.

5. Repeat the Zoombox Process

Maybe two more times and you should be able to trace extremely close to X = 6.

(When I did this, I could trace to

X = 6.0000022, Y = 7.0000023.)

Theanswer is 7.

13. Use your calculator to evaluate

Li—m—3

5 —24

Try all three methods.

Solve /t

1t. Use your calculator to determine

Lim sin—x. Use all three methods.

—0 tan—1

Solve /t

X

Making l/oursetf a Limit Sandwich

The Sandwich Or Squeeze Method is something you can try when you can't solve a limit problem with algebra. The basic idea is to find one function that's always greater than the limit function (at least near the arrow-number) and another function that's always less than the limit function. Both of your new functions must have the same limit as x approaches the arrow-number. Then, because the limit function is "sandwiched" between the other two, like salami between slices of bread, it must have that same limit as well. See Figure 4-1.

§

G

0. What's Lim^= ?

X - "V x

A. The limit is 0.

1. Try plugging in 0. No good, you get 0 over 0.

You should be able to solve this

Limit problem with algebra. But let's say you tried and failed so now

You're going to try the sandwich

Method.

2. Graph the function.

Looks like the limit as x approaches 0 is 0.

3. To prove it, try to find two bread functions that both have a limit of 0 as X Approaches 0.

It's easy to show that the function is always positive (except perhaps at

X = 0) so you can use the simple function § = 0 as the bottom slice of bread. Of course, it's obvious that lim0 = 0. Finding a function for the

X - 0 &

Top slice is harder. But let's just say

That for some mysterious reason,

You know that § = J\x\ Is greater

Near the arrow-number

Than -4=

Y x

The only place that matters for thesandwich method. Because

LimYfxj = 0, § = J\x\ Makes a good

Top slice.

You're done. Because -4= Is

Y x

Squeezed between § = 0 and § = J\x\, both of which have limits of 0 as x approaches 0, -4= Must also have a limit of 0. * x

15. Evaluate l

-0 2

Solve It

16.

Evaluate lim X2cos X

-0

Solve It

Into the Great Beyond: Limits at Infinity

To find a limit at infinity (lim or lim you can use the same techniques from the bulleted list in the "Solving Limits with Algebra" section of this chapter in order to

Change the limit expression so that you can plug in and solve.

If you're taking the limit at infinity of a Rational function (which is one polynomial

Divided by another, such as 5 8x_+ 1_22), the limit will be the same as the §-value

Of the function's Horizontal as§mptote, Which is an imaginary line that a curve gets closer and closer to as it goes right, left, up, or down toward infinity or negative infinity. Here are the two cases where this works:

\/> Case 1: If the degree of the polynomial in the numerator is Less than The degree of the polynomial in the denominator, there's a horizontal asymptote at § = 0 and the limit as X Approaches oo or _ oo is 0 as well.

\S Case 2: If the degrees of the two polynomials are Equal, There's a horizontal

Asymptote at the number you get when you divide the coefficient of the highest power term in the numerator by the coefficient of the highest power term in the

Denominator. This number is the answer to the limit as X Approaches infinity or

Negative infinity. By the way, if the degree of the numerator is greater than the degree of the denominator, there's no horizontal asymptote and no limit.

0. A.

Consider the following four types of expressions: x10, 5x, X!, and xx. If a limit at infinity involves a fraction with one of them over another, you can apply a handy little tip. These four expressions are listed from "smallest" to "biggest." (This isn't a true ordering; it's only for problems of this type; and note that the actual numbers don't matter; they could just as easily be x8, 3x, x!, and xx.) The limit will equal 0 if you have a "smaller" expression over a "bigger" one, and the limit will equal infinity if you have a "bigger" expression over a "smaller" one. And this rule is not affected by coefficients.

For example, lim

1000■ x"

3x!

0 and lim

500■100x

Like (2x)! can change the ordering.

Find lim

X

■ 1.01 The limit is 0.

This is an example of a "small" expression

Over a "big" one, so the answer is 0. Perhaps this result surprises you. You may think

That this fraction will keep getting bigger and bigger because it seems that no matter

What power 1.01 is raised to, it will never grow very large. And, in fact, if you plug

Over

1000 into, the quotient is big

47,000. But if you enter ".0"^ in graphing

Mode and then set both TblStart And Atbl To 1000, the table values show quite convincingly that the limit is 0. By the time x = 3000, the answer is.00293, and when x = 10,000, the answer is 6 x 10"32.

0. A.

Oo. Note, however, that something

Lim

100

5x" cosx2

The limit is 0.

Lim

100

5x" cosx2

100

= ^o~

= 0

The values of cosx2 that oscillate indefinitely between -1 and 1 are insignificant compared with 5x as x approaches infinity.

100

Or consider the fact that lim

5x"10

0

And that

100

100

5x - 10

100

For large

Is always

5x - cosx2

Values of x. Because,. 2 5x - cosx2

Positive for large values of xand less than something whose limit is 0, it must also have a limit of 0.

17. What's lim5x -

10

2x4 + x + 3

? Explain your

Answer.

Solve It

18. What's lim 3x^ answer.

100x3

8x4 + 1

? Explain your

Solve It

X

X — Oo

19. Use your calculator to figure lim ^

Solve It

H ■

20. Determine lim

Solve It

5x+2

4x

1

*2 1. Evaluate lim (4x + /16x2 - 3x)

Solve It

*22. Evaluate lim

Solve It

3x2

3x2

X-1 x+1

Solutions {or Problems ©ith Limits

D Lnn

.....x-i9

X—3 X - 3

Factor, cancel, plug in.

, (x - 3)(x + 3) = lim -

Lim

(x-3)

X + 3 1

3+3

Lim

X-1

1

3

Lim

1x2+x-2

Factor, cancel, plug in.

,. (x - 1)

"m (x - 1)( x + 2)

= lim 1

X — 1 X + 2 = 1

1 + 2 = 1

3

X + 2 1

- — -

-2 X3+

8 = 12

Factor, cancel, plug in.

= lim-(x + 2)-

(x + 2)(x2 - 2x + 4)

1

= lim - 2 _ .

—-2 2-2 +4

1

(-2) - 2 (-2) + 4 12

Lim

2-4

—24 2+5 -6

0

Did you waste your time factoring the numerator and denominator? Gotcha! Always

Plug in first! When you plug 2 into the limit expression, you get or 0 — that's your answer.

D Lim

X - 9

: -6

X—9 3 - /x

1. Multiply numerator and denominator by 3 + FX.

Lim

6

_

2. Multiply out the part of the fraction containing the conjugate pair (the denominator here).

(x - 9)(3 +/x)

= lim - , ,

—9 (9- )

3. Cancel.

=linj (-1 (3+yX))

A - b

Don't forget that any fraction of the form b_a always equals

4. Plug in

= -1 (3 + «/9)

-6

-1.

/x-5-y/5 _ ,/5 Iim x -10 _ 10

Multiply by conjugate, multiply out, cancel, plug in.

= Lim (^-?# ■ t^^5! x—10 (x - 10) (/x - 5 + y5)

Lim

—10

Lim

( -5) -5

(x - 10)7x - 5 + J5)

(x - 10) (x - 10)7x - 5 + J5)

= lim, ,

—10 -5+5

= —f

—i--1—

/10-5

1

2J5

JL

' 10

Cos x - 1

X

0

Did you try multiplying the numerator and denominator by the conjugate of cos x - 1? Gotcha

Again! That method doesn't work here. The answer to this limit is 0, something you just have to

Memorize.

1 - 1

D Lnn

X_2

X—2 X - 2

1

- —4

1. Multiply numerator and denominator by the least common denominator of the little fractions inside the big fraction — that's 2x.

1 - 1

L i—m2

2 2x

2.Multiply out the numerator.

R (2 - x) (x - 2)(2x)

—10

1

3. Cancel.

= Lim XL

X - 2 ZX

4. Plug in.

-1 = T 1

= — —■

Km-.——.— = -36

X - »I + 1 6 x - 6

Multiply by the least common denominator, multiply out, cancel, plug in.

=Limr+zn '!{x-f}

6 x - 6 = Lim 6X (x - 6)

— ( 6) + 6

= ]. 6x(x- 6) = LXImm X

= Lim6 (x- 6)

= 6 (0 - 6)

= -36

M Lim^F = I

-0

No work required — except for the memorization, that is.

Did you get it? If not, try the following hint before you read the solution: This fraction sort of resembles the one in problem 10. Still stuck? Okay, here you go:

1. Multiply numerator and denominator by 3.

You’ve got a 3x in the denominator, so you need 3x in the numerator as well (to make the fraction look more like the one in problem 10).

= Lim ^23- ‘ 3

-0Sin3 3

= Lim„ 3xQ -03sin3

2. Pull the 11 through the Lim Symbol (the 3 in the denominator is really a -, right?).

= 3limsin3x

Now, if your calc teacher lets you, you can just stop here — since it’s "obvious" that limsnnix = 1 — and put down your final answer of 1 – 1, or 3. But if your teacher’s a stickler

For showing work, you’ll have to do a couple more steps.

3. Set u = 3X.

4. Substitute u for 3x. And, because u approaches 0 as x approaches 0, you can substitute u for x under the lim symbol.

1 i;™ «

= 3lin? iIH«

= 7J ■ 1

1

= 3

Because lim^^ = 1, the limit of the reciprocal of, namely, must equal the

Reciprocal of 1 — which is, of course, 1.

E Lim

Tan

1. Use the fact that lini^XP = 1 and replace tanX With.

= Lim’ "

2. Multiply numerator and denominator by cosX.

= lim

= lim

X Cos X Sinr ‘ Cosx

Cosx

Cos

X – O Sinx

3. Rewrite the expression as the product of two functions.

Cos

= lim

Sinx 1

4. Break this into two limits, using the fact that lim(° (x) ■ g (x)) (provided that both limits on the right exist). x -’

Lim ° (x) ■ lim G (x)

^im-iX-x – O Sinx

= 1 ■ 1 = 1

Lim cosx

Limn -

5x – 24

X + 3

-11

You want the limit as x approaches -3, so pick a number really close to -3 like -3.0001, plug that

Into xin your function

5 -24

X+ 3

And enter that into your calculator. (If you’ve got a calcula -

Tor like a Texas Instruments TI-83, TI-86, or TI-89, a good way to do this is to use the STO^

Button to store -3.0001 into x, then enter

5 -24

X+ 3

Into the home screen and punch Enter.)

The calculator’s answer is -11.0001. Because this is near the round number -11, your answer is -11. By the way, you can do this problem easily with algebra as well.

Sin

1

Sin

Then go to Table setup And enter a

0 tan

Enter the function in graphing mode like this: § , .

Tan ^

Small increment into Atbl (try 0.01 for this problem), and enter the arrow-number, 0, into TblStart. When you scroll through the table near X = 0, you’ll see the Y Values getting closer and closer to the round number 1. That’s your answer. This problem, unlike problem 13, is not easy

To do with algebra.

X

1

_

—

± Evaluate lim (X Sin-k^

Here are three ways to do this. First, common sense should tell you that this limit equals 0. lim x is 0, of course, and lim (sin -X) never gets bigger than 1 or smaller than -1. You could say that,

Therefore, lim (sin Xr) Is "bounded" (bounded by -1 and 1). Then, because Zero X Bounded = Zero,

X — O y X I

The limit is 0. Don’t try this logic with you calc teacher — he won’t like it.

Second, you can use your calculator: Store something small like.1 into x and then input x sin X.

X

Into your home screen and hit enter. You should get a result of ~ - .05. Now store 0.01 into x

And use the Entry Button to get back to xsin-^ and hit Enter Again. The result is ~ .003. Now

X.

Try.001, then.0001 (giving you ~ - .00035 and ~ .00009) and so on. It’s pretty clear — though

Probably not to the satisfaction of your professor — that the limit is 0.

The third way will definitely satisfy those typically persnickety professors. You’ve got to

Sandwich (or Squeeze) Your Salami Function, xsinXr, between two Bread Functions that have

X.

Identical limits as X Approaches the same arrow-number it approaches in the salami function.

Because sinX – never gets bigger than 1 or smaller than -1, XSin X – will never get bigger than

X. x. \x\ Or smaller than -X|. (You need the absolute value bars, by the way, to take care of negative values of x.) This suggests that you can use B (x) =-|x\ For the bottom piece of bread and

T (x) = \x\As the top piece of bread. Graph B (x) = -|x|, F (x) = xSin, and T (x) = \x\At the same time on your graphing calculator and you can see that XSin – X is always greater than or equal to -X\And always less than or equal to |x|. Because lim(-X|) = 0 and lim|x| = 0 and

Because XSinX – is sandwiched between them, lim(XSin-^) must also be 0. ° Evaluate lim(X2cos-) = 0

—0

For lim(X2cos1 J, use B (x) = – x2 And T (x) = x2 For the bread functions. The cosine of anything

Is always between -1 and 1, so x2cos:1 is sandwiched between those two bread functions. (You should confirm this by looking at their graphs; use the following window on your graphing calculator — Radian Mode, XMin = -0.15625, XMax = 0.15625, XScl = 0.05, YMin = -0.0125, YMax =

0.0125, YScl = 0.05.) Because lim(-X2• = 0 and limX2 = 0, lim(X2cos-x) is also 0.

—0 —0 —0

M Lim5x 3~x 2130=0

X — =° 2x +x + 3

Because the degree of the numerator is less than the degree of the denominator, this is a Case 1 problem. So the limit as X Approaches infinity is 0.

® lim

3×4 + 100×3 + 4 = 3 8×4 + 1 8

Lim 3×4 +100×3 + 4 is a Case 2 example because the degrees of the numerator and denominator are both 4. The limit is thus 3.

EU Lim §7 = -

According to the "big" over "small" tip, this answer must be infinity. Or you can get this result with your calculator. If you set the table (don’t forget: fork on the left, spoon on the right) the same as in problem 19, you’ll see "undef" for all Y Values. You’ve got to be careful when trying to interpret what "undef" (for "undefined") means on your calculator, but in this case — because a 0 denominator is impossible and thus can’t be the reason the fraction is undefined — it looks like "undef" means infinity. To confirm this, make TblStart And Atbl Smaller, say, 10. Sure enough, they values grow huge very fast, and you can safely conclude that the limit is infinity.

— lim

5X + 2

1

1. Divide numerator and denominator by x.

5X + 2

= Lim -

4

■x

1

X

2. Put the x into the square root (it becomes x2).

5X + 2

= Lim -

■ /4X2 - 1

Ґ x2

3.Distribute the division.

5 + x = Lim-X

—

2

4. Plug in and simplify.

2

Y4 -

5+0

/4 - 0

= 5 2

*« Hm (4x +/16xT-ax) = 3

1. Put the entire expression over 1 so you can use the conjugate trick.

(4x + / 16×2 - 3x) (4x -/ 16×2 - 3x)

Lim

1 (4x - /16×2 - 3x)

2. FOIL the numerator.

16×2 - (16×2 - 3x)

Lim

4x -/16×2 - 3x

3. Simplify the numerator and factor 16×2 inside the radicand.

Lim

3

VM1 – ilx

4

4. Pull the 16×2 out of the square root — it becomes -4x.

It becomes Negative 4X Because X Is negative when X— - -, and because you’ve got to pull out a Positive, You pull out -4x because when X Is negative, -4X Is positive. Got it?

= lim

3

4x – (-4x )f7Ix

Lim

3

4X (1

Y116x

5. Cancel and plug in.

Lim

3

4 (1

A7*)

1-1

6(- )

=-7

4 (1 + /T-0)

-% Piece o’cake.

Lim I -3×4- – -3xi) = 6

(x – 1 x+ 1,

1. Subtract the fractions using the LCD of (x – 1)(x + 1) = x2 – 1.

Lim3x2(x + 1 y2(x - 1) — 2-1

2.Simplify.

3 3+3 2-3 3+3 2

Lim

Lim

2-1

6 2

— 2-1

3. Your answer is the quotient of the coefficients of x2 in the numerator and the denominator (see Case 2 in the "Into the Great Beyond" section).

= 6

Note that had you plugged in in the original problem, you would have

-1 +1

=— - — 0?

It may seem strange, but infinity minus infinity does Not Equal 0.

3

3