Calculus Sin Frontera · 2020-06-27 · calculus sin frontera 1 Section 1.1: Removable Discontinuities This mini-section is about doing limits for removable discontinuities. we actually

calculus sin frontera 1

Section 1.1: Removable Discontinuities This mini-section is aboutdoing limits for removable discontinuities. we actually have four dif-ferent kinds of things to think about: intuitions, definitions, graphicalexamples, and last, computational techniques.

Intuitions My quick and dirty guide is that f has a removable dis-continuity at c if you plug c into f and you get a 0/0. That’s indeter-minate (see the ’Dividing by Zero’ mini-section) so I have to do somealgebra to get a definite limit. The algebra usually means cancelingsomething from the numerator and from the denominator. I cancelout the zero from each. what’s left should give an actual number,when you plug back c back in.

Definition of Removable Discontinuity The function y = f (x)has a removable discontinuity at x = c if f (c) does not exist, butlimx→c f (x) exists and is finite.

This incorporates the idea that plugging c into f doesn’t work, butI can still cancel out the zeroes. Now I need a theorem that lets mecancel, but still keeps the same limit. The guy with no cancellingwe’re calling f ; the guy with the cancelled zeroes is g:

Theorem Assume f and g are functions, and f (x) = g(x) for allx 6= c. Then limx→c f (x) = limx→c g(x)

Good: I can cancel and still keep the limit. Next, what to cancel to getrid of the zeroes:

Theorem Assume P(x) is a polynomial in any mixture of x, 1x ,√

x.Assume P(c) = 0. Then P has a factor of (x− c).

Now were ready: we know when to factor, we know what to factor,and we know the factoring will work. It’s time for:

Worked Problems

Problem limx→2

x3 − 2x2 + x− 2x3 + x2 − 4x− 4

Step 1 Plug in c, that is, x = 2 (you get c from whatever is after thearrow in x →).

f (2) =23 − 2 · 22 + 2− 223 + 22 − 4 · 2− 4

=8− 8 + 2− 28 + 4− 8− 4

=00

2 Kathy Davis

Figure 1: The Graph:Here’s the graph; note the circle atx = 2. We knew the circle was at x = 2,because that was in the limit. When wetook the limit, what we were computingwas the y = 5/12 height of the circle.5/12 ≈ .42 but on this graph the bestyou can see is y is close to .5

Step 2 Now to factor x− c = x− 2 from numerator and denominator.When I do this, I’m gonna be sneaky: In the numerator i see a termlike 8-8; it came from 23 − 2 · 22, which came from x3 − 2x2. Since8− 8 = 0, that means the term x3 − 2x2 already has a factor of x− 2,all by itself. And it does: x3 − 2x2 = x2(x− 2).

Thinking along these lines, the numerator is x3 − 2x2 + x − 2 =

(x3 − 2x2) + (x− 2) = x2(x− 2) + (x− 2)(1) = (x2 + 1)(x− 2)

The denominator was 8+4-8-4. I need to rearrange it to get 8-8+4-4.But 8+4-8-4 came from x3 + x2 − 4x− 4; when I re-arrange that, I getx3 − 4x + x2 − 4 = (x3 − 4x) + (x2 − 4) = x(x2 − 4) + (x2 − 4) =

(x + 1)(x2 − 4) = (x + 1)(x + 2)(x− 2).

Step 3 Cancel:

x3 − 2x2 + x− 2x3 + x2 − 4x− 4

=(x2 + 1)(x− 2)

(x + 1)(x + 2)(x− 2)=

(x2 + 1)(x + 1)(x + 2)

Step 4 Take the limit:

limx→2

x3 − 2x2 + x− 2x3 + x2 − 4x− 4

= limx→2

(x2 + 1)(x + 1)(x + 2)

=(4 + 1)(3)(4)

= 5/12

Remark 1 If the whole 8+4-8-4 business drove you crazy, you couldhave also just done a long division: divide numerator and denomina-tor each by x − 2 (don’t try to long divide the denominator into thenumerator; it’s the x− 2 we’re after).

Remark 2 Notice how sneaky I was, doing all my algebra separately,then plugging into the limit at the last step. That way, I didn’t have towrite limx→2 every single step. And yes, I do take off points for that.

Remark 3 Wasn’t there supposed to be a ’graphical examples’ here,somewhere? Sigh. Notice the two strong green lines; they’re asymp-totes, which come from when the denominator, (x + 1)(x + 2), iszero.

This would have been a good time to use L’Hospital’s rule!


Problem limx→−1

3−√

8− x1 + x

Step 1 Plug in c, that is, x = −1.

f (−1) =3−√

8−−11 +−1

=3−√

8 +−11− 1

==3−√

90

=00

Step 2 Now to factor x − c = x −−1 = x + 1 from numerator anddenominator. Denominator is easy: 1 + x = x + 1. It’s the numeratorthat worries me. I can’t see an x + 1 in 3−

√8− x to save my life.

One thing’s certain: I need to get rid of the square root. Just likethe last problem, there’s a sneaky way to do this. Say you have, asI do, a − √y. Notice if I multiply this by the conjugate, a +

√y, I

get (a − √y)(a +√

y). But this is a difference of squares, so I get(a)2 − (

√y)2 which is a2 − y. No more square roots. Of course, I

can’t go around multiplying by a +√

y whenever I feel like it. I canmultiply by a factor of one, though:

3−√

8− x1 + x

=(3−

√8− x)

(1 + x)· (3 +

√8− x)

(3 +√

8− x)=

(3−√

8− x) · (3 +√

8− x)(1 + x) · (3 +

√8− x)

Now use the difference of squares trick in the numerator to get

=32 − (

√8− x)2

(1 + x) · (3 +√

8− x)=

9− (8− x)(1 + x) · (3 +

√8− x)

=(1 + x)

(1 + x) · (3 +√

8− x)=

13 +√

8− x

It’s limit time!

limx→−1

3−√

8− x1 + x

= limx→−1

13 +√

8− x=

13 +√

8−−1=

13 +√

8 + 1=

13 +√

9=

13 + 3

=16

4 Kathy Davis

Section 1.2: Long Division

Let’s Long! This is for all you who forgot how to long divide polyno-mials. I’ll start with a simple case:

f (x) =x3 + x2 + x + 1

x2 − x− 1

Here’s the entire division; in a bit, we’ll do it one step at a time:

x + 2

x2 − x− 1)

x3 + x2 + x + 1− x3 + x2 + x

2x2 + 2x + 1− 2x2 + 2x + 2

4x + 3

What this means is that x2 − x− 1 divides into x3 + x2 + x + 1 exactlyx + 2 times, with a remainder of 4x + 3:

f (x) =x3 + x2 + x + 1

x2 − x− 1= (x + 2) +

4x + 3x2 − x− 1

Now I’ll do the division step-by-step: first, set up the division:

x2 − x− 1)

x3 + x2 + x + 1

Next, look at the highest powers – a x3 in numerator and a x2 indenominator, and ask, ’how many times does x2 go into x3? x times,of course, so you write that on the top of the division:

x

x2 − x− 1)

x3 + x2 + x + 1

now you take that x and multiply it into the x2 − x − 1, to get x3 −x2 − x. Put that under the original x3 + x2 + x + 1 in the division,multiply by -1 (because we’re going to subtract) to get

x

x2 − x− 1)

x3 + x2 + x + 1− x3 + x2 + x

Now add the two lines, to do the subtraction:

x

x2 − x− 1)

x3 + x2 + x + 1− x3 + x2 + x

2x2 + 2x + 1


Now you start the whole process all over! You’ll divide x2 − x − 1into 2x2 + 2x + 1. The x2 foes into the 2x2 twice, so put that on top asa +2:

x + 2

x2 − x− 1)

x3 + x2 + x + 1− x3 + x2 + x

2x2 + 2x + 1

Now multiply the 2 into the x2 − x− 1 and subtract so that you get a−2x2 + 2x + 2. Put that under the 2x2 + 2x + 1

x + 2

x2 − x− 1)

x3 + x2 + x + 1− x3 + x2 + x

2x2 + 2x + 1− 2x2 + 2x + 2

Now add:x + 2

x2 − x− 1)

x3 + x2 + x + 1− x3 + x2 + x

2x2 + 2x + 1− 2x2 + 2x + 2

4x + 3

Now try to divide x2 − x− 1 into 4x + 3. x2 doesn’t go into 2x, so thedivision stops here and 4x + 3 is the remainder.

Here’s a couple more; try them yourself, then check the next page:

f (x) =x4 − 2x + 2x− 3

x− 2f (x) =

x3 − x2 − xx2 + x− 3

6 Kathy Davis

x3 + 2x2 + 4x + 8

x− 2)

x4 − 3− x4 + 2x3

2x3

− 2x3 + 4x2

4x2

− 4x2 + 8x

8x − 3− 8x + 16

13

x− 2

x2 + x− 3)

x3 − x2 − x− x3 − x2 + 3x

− 2x2 + 2x2x2 + 2x− 6

4x− 6


Section 1.3: Division By Zero When you start talking about thesethings, it’s easy to get into long pointless arguments: ’it should bethis.”No, it has to be that.’ Like asking ’can Wonder Woman beatSuper Girl?’ It can go forever.

In society, we resolve those kinds of disputes with laws and trials andarbitration boards. These are supposed to set the societal standardswhere we appeal for justice. In math, we do it differently: the lawsand standards are definitions and theorems.

Definition

ab= c means a = bc

If I say something like 1/0 = 2. The Law says that what I must reallymean by that is 1 = 2 · 0. So I’m saying 1 = 0. Sure it is.

This what a definition does for me: it clears away a lot of words andgives me something to calculate, to check. And I see right away that1/0 = x is never gonna work, no matter what x I try. So: 1/0 does notexist. And we mean there’s no x which would be equal to 1/0. We’llabbreviate ’does not exist’ as dne.

Now let’s try computing 00 . Someone could say, ’any number divided

by itself is one’ and so 00 = 1. Or someone else could say, ’ 0

0 = 01 · (

00 )

and any number times zero is zero.’ So, 00 = 0.

Oh great: already we have two different answers. The point still is,we can say lots of things, but how do we calculate? We appeal to thelaw: the definition.

00= 1 means 0 = 0 · 1 true!

00= 0 means 0 = 0 · 0 true!

00= π means 0 = 0 · π true!

They can’t all be right! When we compute we’re supposed to get ananswer. One answer. So now we have something different: 0/0 couldbe equal to anything. This time we’ll say: 0/0 is indeterminate.

We agree on two standards:10 does not exist.00 is indeterminate.

What we do not do is assign a number to either of these.

8 Kathy Davis

Section 1.4: Why Can’t We Use Calculators? Two reasons:

One, these computations are the kinds of skills needed for the restof the course. A lot of this course isn’t about just getting answers;you can do that faster using Google or Wolfram Alpha. A lot of thecourse is about learning techniques, also when and where to applythose techniques.

Two: calculators have some issues with accuracy. They can be likeasking a friend who is really fast but also likes to play tricks on you.You never know ....

Here’s an example:

limx→0+

1− cos(x)x2

So OK, let’s do this with a calculator: we need to let x get nearer andnearer to zero, so we’ll let x be .1, .01, .001 and, bored! writing zeroes,so we’ll write 10−4, 10−5, 10−6, . . .. Remember the idea of a limit(even with a calculator): limx→c f (x) = L means the more decimalplaces x and c agree, the more decimal places f (x) and L agree. Ifwe’re taking x = 10−n, the larger n is, the better answer we’ll get forthe limit. Ready?

x (1− cos x)/x2

.1 0.45970

.01 0.50000

.001 0.50000

10−50.50000

10−60.50004

10−70.49960

10−80

10−90

Figure 2: Xena Says:What do you mean, I can’t use mycalculator?

Well, that’s it, then, huh? The big n win, the answer is zero, not.50000. All good, except, it’s wrong:

1− cos(x)x2 =

1− cos(x)x2

1 + cos(x)1 + cos(x)

=1− cos2(x)

x21

1 + cos(x)

=sin2(x)

x21

1 + cos(x)=

(sin(x)

x

)2 11 + cos(x)

So now that’s over, let’s take the limit:

limx→0+

1− cos(x)x2 = lim

x→0+

(sin(x)

x

)2 11 + cos(x)

= (1)2 11 + cos(0)

=12


!So the .5 was right, and the zero is wrong? That can’t be! I mean,calculators are supposed to be accurate to like 10 or 12 decimalplaces! And here the calculator is screwing up at 10−9 which is onlylike nine decimal places?

Relax: it isn’t the fault of the calculator – at least not very directly. It’sa decimal place thing. Calculators can only carry so many decimalplaces. If two numbers a, b look the same to ten or twelve decimalplaces, then when you compute a− b the calculator will only see theplaces where they’re the same, and will give you a zero.

In this case, cos(10−9) = 0.9999999999999999995. If you round this,correctly, to 16 decimal places, it’s a 1. So the calculator thinks1− cos(10−9) = 0.

Some students say, "I’m gonna be an engineer, and we do everythingwith computers, so I’ll never use this stuff." Well, OK.But, here’s agreat little line; I copied it from my calculator instruction book:

In no event shall Texas Instruments be liable to anyone for special, collat-eral, incidental or consequential damages in connection with arising outof the purchase or use of this calculator or manual.

If we use a calculator or computer to do a computation, and thecalculator has a bug, we’re screwed: we can’t sue. For everythingelse, there’s limits.

This is actually kind of bad: can I ever trust a calculator or a com-puter? Regaining trust is one of the things math can do. The tech-nical term for what just happened is ’loss of significance’ and a websearch on that phrase gives a lot more about why and when thingsgo wrong. The area of mathematics that studies issues like this is Nu-merical Analysis. So there’s a theory of mistakes, and how to avoidmistakes.

Short of taking an entire extra course, though, there’s a simple rule:when we do real scientific and engineering computations, we shouldtry to avoid subtracting two numbers that are almost equal. Rear-range subtractions into additions. If we let

1− cos(x)x2 =

1− cos(x)x2

1 + cos(x)1 + cos(x)

=1− cos2(x)

x2 =sin2(x)

x21

1 + cos(x)

and plug x = 10−9 into that, we get a lot better answer: 0.499999999999999999958.Which rounds to .5.

Of course, it would’t be right to leave without some examples:

limx→0

x− sin(x)x3 lim

x→0

1−√

1 + x2

x2 limx→0

1− (1 + x2)5

x8

10 Kathy Davis

If you know any Taylor Series, you can use the series to help withthese kinds of computations. For example, try using the Taylor Seriessin x ≈ x− x3

3! to do the first limit.


Section 1.5: Hello? L’Hospital’s Rule? I was online with a highschool AP teacher a few years ago. Call him Mr. Sagredo. He asked,"What is wrong with you people at UT? You’d help your students alot if you let ’em use L’Hospital’s Rule. I mean, look at what you do:"

1) limx→2x3−2x2+x−2x3+x2−4x−4 long division

2) limx→1x2−11−√

x rationalize the square root

3) limx→11− 1

x1−√

x rationalize; common denominator

4) limx→0sin x

x squashing theorem

"Four different limits and each one needs some different kind ofalgebra. What a mess! With L’Hospital’s rule, you’d just differentiate’em all."

But calculus at UT isn’t about getting answers; it’s about learningdifferent techniques, and when to use each. It’s the same in everyscience, craft or profession. There’s no wonder drug that cures ev-ery disease; there’s no one surgical technique you can use on everyorgan. If you want to design a building there’s no one constructionmaterial you can put everywhere; you have to know what to putwhere. And in calculus, there no magic formula that does every limit:you have to know what technique to use, and when to use it.

Here’s some problems: try using only L’Hospital on each (no algebra,now). It isn’t gonna happen.

limx→∞

x√x2 + 1

limx→0+

x ln(x)

12 Kathy Davis

Figure 3: Zeroes and signs of x3 − x.The green holes are where x3 − x = 0;In blue, x3 − x > 0; in red, x3 − x < 0.

Figure 4: signum(x3 − x).The green holes are where x3 − x = 0;these show where the jump discontinu-ities happen.

Section 1.6: Jump Discontinuities We’ve seen how to deal withlimits for piecewise-defined functions, for example:

f (x) =

x + 1 if x > 1

x2 if x <1

dne if x=1

=

gR(x) if x > 1

gL(x) if x <1

dne if x=1

Thenlim

x→1+f (x) = lim

x→1gR(x) = lim

x→1(x + 1) = 2

andlim

x→1−f (x) = lim

x→1gL(x) = lim

x→1(x2) = 1

Since limx→1+ f (x) 6= limx→1− f (x), we say that limx→1 f (x) does notexist, and f has a jump discontinuity at x = 1.

As in all these limits, the key is to reduce everything to a small num-ber of basic types, that we know how to compute. There are threerules.

Rule One The signum function is defined as

signum(x) =x|x|

Then for any function f (x),

signum[ f (x)] =f (x)| f (x)| =

1 if f(x) > 0

−1 if f(x) < 0

dne if f(x) = 0

Another way to say it is that signum asks a function, ’hey, what’syour sign?’ And the function answers +1 if it’s positive, and -1 if it’snegative. Here’s an example: f (x) = x3 − x. To see where f changesfrom + to -, I’ll first check where it’s zero. f (x) = x3 − x = x(x2 −1) = x(x− 1)(x + 1). The whole point of factoring is, f is zero whenthe factors are zero. So, f is zero when x = 0, x = 1, x = −1. Andhere’s f , with its positive and negative parts, in Figure ??. Notice thegreen circles.

Now here’s signum(x3 − x), in Figure ??. The signum function canonly take on vales of +1 and -1, which we see as the heavy blacklines. signum switches from + to - at the green circles – which matchthe green circles in Figure ??.


Rule Two We need to know a little more about the signum function.Here’s something really useful:

signum(x) =1

signum(x)

The easiest way to see this is to think what it means: signum is either+1 or -1. If you take one over +1, you still get +1; ditto for -1.

Where you see this is in problems like – well, |x3 − x|/(x3 − x)

|x3 − x|(x3 − x)

=1

x3−x|x3−x|

=1

signum(x3 − x)= signum(x3 − x) =

x3 − x|x3 − x|

Rule Three We also need to know a little about the absolute valuefunction. Here it is: |x · y| = |x| · |y|. Where we see this is in problemslike

|x3 − x| = |x(x− 1)(x + 1)| = |x| · |x− 1| · |x + 1|

.

Alright; it’s about time we actually did a problem or two. Let’s take

limx→−1

|x2 − 1|x2 + x

When you plug in -1, you get |(−1)2−1|(−1)2−1 = |1−1|

1−1 = |0|0 . That’s the key

to use left and right limits. Before I can do that, though, I have toget my gl , gR. And I get those from signum. Where’s signum? First,I want to think: what signum do I want? That’s the easy part: I havelimx→−1 so, I want signum(x−−1) = signum(x + 1). Here goes:

|x2 − 1|x2 + x

=|x− 1| · |x + 1|

x(x + 1)=|x− 1|

x· |x + 1|

x + 1=|x− 1|

x· signum(x+ 1)

|x− 1|x· signum(x + 1) =

|x− 1|x·

1 if x > -1−1 if x <-1dne if x=-1

=

|x−1|

x if x > -1

− |x−1|x if x <-1

dne if x=-1

=

gR(x) if x > -1gL(x) if x <-1dne if x=-1

Now let’s do the limits:

limx→−1+

f (x) = limx→−1

gR(x) = limx→−1

|x− 1|x

=| − 1− 1|−1

= −2

14 Kathy Davis

and

limx→−1−

f (x) = limx→−1

gL(x) = limx→−1

−|x− 1|x

= −| − 1− 1|−1

= 2

Figure 5: A Limit Example.

Graph of |x2−1|

x2+x . The green holes are atthe jump, x = −1.

−2 6= 2 so limx→−1 f (x) does not exist. If you look at Figure ??, yousee the jump.

Remark This example is different from our usual ones: here the leftlimit is positive, and the right limit is negative. The factor of 1/x isnegative at x = −1, and this causes the limits to be different in signthat we’re used to.

There’s a moral: you can’t take anything for granted: you have tocompute the limits.


Section 1.7: Weird Limits Level II Alert This one goes out to allthose strange limits that no-one loves. We’ll start with limx→∞ sin(x).A little graph shows sine doesn’t look remotely like it has a limit – itwobbles between +1 and -1 and never settles down.

Figure 6: sin(x)Looking like it’ll wobble forever.

Figure 7: sin(x)/xTaming the wobble.

Figure 8: A Close Up

Here I can see sine being squasheddown by the green ±1/x.

But Let’s try it: let’s pretend I believe limx→∞ sin(x) = 1. What’sgonna happen? I say I believe that sine settles down near one. Butthe very next cycle, sine hits zero: sine refuses to stay close to just onenumber.

Now let’s make that intuition into math (algebra, actually): I’m sup-posed to believe that there’s a number n, and when x > 10n, sin(x)and 1 agree to, say, two decimal places. Since sine can’t be biggerthan 1, sin(x) would have to be between 1 and, say, .95: for x > 10n,.95 ≤ sin(x) ≤ 1. That means if I take x = 10nπ, then I certainly havex > 10n. So I’d have to believe that .95 ≤ sin(10nπ) ≤ 1. But sine iszero at all multiples of π, and if I use that, .95 ≤ 0 ≤ 1. So not gonnahappen.

My next limit is a variant: limx→∞ sin(x)/x. The idea here is thatthe sine wobbles, but the 1/x goes to zero, so that multiplying the1/x into sine will tame the wobble. Like in Figure ??: the wobble istamed.

Here it is in math: The wobble is

−1 ≤ sin(x) ≤ 1

And the taming is

− 1x≤ sin(x)

x≤ 1

x

Now for the limit:

limx→∞

− 1x≤ lim

x→∞

sin(x)x≤ lim

x→∞

1x

And now

0 ≤ limx→∞

sin(x)x≤ 0

As promised, all the wiggle has been squashed out of sine; it has nochoice but to go to zero.

Figure 9: Flipping The Sine

The wiggles of sine are compressednear zero.

A Neat Trick This trick interchanges zero and infinity. It’s easy: if Ilet y = 1/x, then as x → ∞, y → 0. I’ll use this to change variables inlimy→0 sin( 1

y ): the limy→0 ↔ limx→∞; the sin( 1y )↔ sin(x), and so:

limy→0

sin(1y) = lim

x→∞sin(x) = does not exist

16 Kathy Davis

I get two limits for the work of one. All the wiggling I saw in Figure?? as x went to infinity, is switched over to zero, and compresseddown; as in Figure ??.

Figure 10: Squashing The Sine, IIThe wiggles of sine near zero aresquashed between the two green lines.

And, let’s do it again, with limy→0 y sin( 1y ). The only term that’s

different is the factor of y in front of the sine: since y = 1/x, we get

y sin(1y)↔ sin(x)

x

and then

limy→0

y sin(1y) = lim

x→∞

sin(x)x

= 0

You can see the squashing in Figure ??.

Figure 11: The Old Switcheroo

When I switch the limit as x → ∞ ofln(x), I create a vertical asymptote forlog at x = 0+.

This trick, switching zero and infinity, works for other functions too.In class, we showed limx→∞ ln(x) = ∞. Now I’m going to switch thatinfinity to zero: let x = 1/y. Then,

limy→0+

↔ limx→∞

ln(x)↔ ln(1y) = − ln(y)

limx→∞

ln(x) = ∞↔ limy→0+

− ln(y) = ∞

So,lim

y→0+ln(y) = −∞

This means that the log function will have a vertical asymptote at atx = 0+, as in Figure ??.

Level III Alert This trick of changing one variable into anotheris something we’ll use repeatedly in the course, in the chapters onintegration. But – here’s an example you might not expect:

limx→1

1− 3√

xx− 1

=00

Since it’s a 0/0, I want to factor out an x − 1 from the numerator. Ifthe numerator had a square root instead of a cube root, I’d multiplyby the conjugate. But what’s the conjugate for a cube root?


Since I don’t know the answer to that (secretly I do [show-off!]), I’lljust get rid of the cube root: let x = y3. Then:

limx→1↔ lim

y→1

1− 3√

xx− 1

↔ 1− yy3 − 1

Difference of cubes, I can handle:

1− yy3 − 1

=−1(y− 1)

(y− 1)(y2 + y + 1)=

−1y2 + y + 1

and,

limx→1

1− 3√

xx− 1

= limy→1

−1y2 + y + 1

= −13

By the way, if you take the 1/(y2 + y + 1) and change back to x, youget

1

x23 + x

13 + 1

and this tells you that the conjugate of 1− 3√

x is x23 + x

13 + 1

:) :) Sleep tight ...

18 Kathy Davis

Section 1.8: The Variety of Asymptotic Experiences

We talked in class about the function y = 1/x when x is large.Whether you do it with a table

x-value y = 1/x

10 0.1100 .01

1000 .001

Figure 12: Global: large xWhen x gets large, y = 1/x gets small.

or with a graph (see Figure ??), the result is the same: we writelimx→∞

1x = 0, or we say ’the line y = 0 is a horizontal asymp-

tote.’ This is fine; but we’re going to find is that it’s too limited: thereare all kinds of other asymptotes, and ’vertical asymptote - horizontalasymptote’ gives only a few of all the asymptotes there are. We’realso talk about our asymptotes differently: instead of saying "the liney = 0 is a horizontal asymptote of y = 1/x", we’ll say "y = 1/x isasymptotic to y = 0."

Definition: We say the function y = f (x) is asymptotic to the functiony = g(x) if limx→∞ [ f (x)− g(x)] = 0.

That is, f and g are asymptotic if they get infinitely close as x getslarge. (same thing works if x → −∞).

Here’s another example, one we know from the Local/Global

section: y = x/(x2 + 1). In that section, we said x/(x2 + 1) ≈ 1/x andwe checked this with charts and graphs. Now, we’ll say x/(x2 + 1)is asymptotic to 1/x, and instead of using a chart or a graph, we’llcheck this algebraically:

limx→∞

[x

x2 + 1− 1

x

]= lim

x→∞

[x2

x(x2 + 1)− x2 + 1

x(x2 + 1)

]

= limx→∞

−1x(x2 + 1)

=−1∞

= 0

You can see this graphically in Figure ??.

Figure 13: Asymptotic To

y = x/(x2 + 1) is asymptotic toy = 1/x. The graph tells you a littlemore: it looks like 1/x is always ontop of x/(x2 + 1). Can you tell why, bylooking at the limit computation?

Figure 14: A Mixed Problem

y = x + 1/x will have two asymptoticbehaviors.

For my next example, I want to mix the ideas from Local/Global

with those from ’asymptotic to’. The example is y = x + 1/x; checkout Figure ??. Here ’local’ means x near zero; try x = .1 for example:y = x + 1

x = .1 + 1.1 = .1 + 10

Local/Global thinking tells me the ’10’ is the important part ofthe function, and the ’.1’ unimportant. The ’10’ comes from the 1/xpart of y = x + 1/x; this means that near zero, y = x + 1/x isapproximately like y = 1/x.


Figure 15: Mixed Problem Near x = 0y = x + 1/x is like y = 1/x near zero.

The two are graphed together in Figure ??. You can see that nearzero, the two are similar, but as x gets larger, another curve takesover. This is the ’global’ part; let’s try x = 10 to count for ’large x’:

y = x +1x= 10 +

110

= 10 + .1

Figure 16: Mixed Problem Near x = ∞y = x + 1/x is like y = x near infinity.

Again the ’10’ is the important part of the function, and the ’.1’ unim-portant, but this time the ’10’ comes from the x part of y = x + 1/x.So for large x, y = x + 1/x is approximately like y = x. Let’s add thatto the graph; you can see the global part in Figure ??.

Remember, though: while we’re not gonna change our ideas abouthow to get ’local’, how we do ’global’ has been changed: the politi-cally correct way to talk is now ’y = x + 1/x is asymptotic to y = x’.And we don’t rely on plugging in ’10’ or intuitions like that: we’rescientists and egnineers, and we compute things. Which, here, isn’texactly hard:

limx→∞

[ f (x)− g(x)] = limx→∞

[(x +

1x

)− (x)

]= lim

x→∞

[(1x

)]= 0

OK, y = x + 1/x is asymptotic to y = x after all. Now let’s put it alltogether:

Figure 17: All Together

y = x + 1/x is caught between twoasymptotes.

Figure ?? shows how the graph of y = x + 1/x sits between thetwo asymptotes. You’ll notice the picture shows that y = x + 1/x isalways ontop of its asymptotes. That’s just algebra: when we graph,we’re only looking at x > 0. Then y = x + 1/x > x so y = x + 1/x isontop of the straight line, and, y = x + 1/x > 1/x so y = x + 1/x isontop of the vertical asymptote.

Here’s a couple of examples for you to think about:a) What does y = x2 + 1/x look like? What it is asymptotic to?b) Same question, but with y = x2 + 1/x2.c) Same question, but with y = x + 1/x2.

We’re not through yet; there are still more funny asymptotes. Exam-ples a) and b) were pretty easy; you were handed the ’asymptotic to’part by writing y = x2 + something. For the next problem, we’ll makeit a little harder to see what each piece is.

20 Kathy Davis

Figure 18: A Hard Example

No vertical asymptotes; any others?

Our new function is y = f (x) = x3/(x2 + x + 1). If you completethe square, you see x2 + x + 1 = (x + 1

2 )2 + 3

4 , which means the de-nominator is never zero, which means f (x) doesn’t have any verticalasymptotes. You see that in Figure ??, Next question: is it asymptoticto anything? How would I even figure that out?

Figure 19: Try Guessing?That actually doesn’t work very well.

One thing to try is using that old high-school trick they tell you:you can get the limits by taking the highest power in the numerator(that’d be x3) and the highest power in the denominator (x2). thenthe whole function looks like x3/x2 = x. Try it; Figure ?? gives youan idea. Also, the idea it gives you is that the two aren’t asymptotic;they keep their distance from each other.

Oh well: good-bye high school. What we want is something likex2 + 1/x. Can we write our f (x) as

f (x) = g(x) + r(x)?

If we could, we’d get

limx→∞

[ f (x)− g(x)] = limx→∞

[r(x)]

and all we’d need is for r(x) → 0, to get f asymptotic to g. Whateverg is.

Figure 20: Long Division!f is actually asymptotic to g(x) = x− 1,not to g(x) = x.

The trick here is long division. Say I let x = 10; then f (10) is 1000111 .

I want to write that as a big number plus a smaller term, and that’swhat a long division does: 1000

111 = 9 + 1111 . In f , g terms, if you do the

long division, you get

f (x) =x3

x2 + x + 1= (x− 1) +

1x2 + x + 1

= g(x) + r(x)

just as I wanted. And, because 1/(x2 + x + 1) → 1/∞ = 0, we getx3/(x2 + x + 1) is asymptotic to x− 1. Just like in Figure ??.

Figure ?? also tells you a little more: the curve lies ontop of theasymptote. Again, there’s an algebraic reason for that:

f (x) = x3/(x2 + x + 1) = (x− 1) +1

x2 + x + 1> x− 1 = g(x)

So we see graphically and algebraically that f > g, so f is ontop of g.


Vertical asymptotes represent places where the function doesn’t exist,so a function cannot "cross" its vertical asymptote. We’ve seen twokinds of asymptotes here, where again the function doesn’t crossthe asymptote. But that’s just because we haven’t done very manyexamples; here’s a simple one.

Figure 21: f Crosses Its Asymptote

How would you figure out where thatcan happen?

f (x) =x2 + x

x2 − x + 2

First you’d need to long divide, and see, as in Figure ??, that g(x) =

1. You’d need to compute f − g and then compute thatlimx→∞ [ f (x)− g(x)] = 0. And then you’d probably want to figureour why, and where, f crosses g.

Of course, you don’t get out of here that easily:a) What does (x3 − x2)/(x2 + 1) look like? What it is asymptotic to?b) Same question, but with y = x3/(x− 1).c) This is evil: y =

√(x2 − 2x)/(x− 3).

Remark What is this? Why does this work? Why long division?

When I write

f (x) =x3

(x2 + x + 1= (x− 1) +

1x2 + x + 1

= g(x) + r(x)

what I’m doing is writing f as two parts: the g(x) contains all theparts of f that don’t go to zero at infinity, and the r(x) all the parts off that do go to zero at infinity. Why? Because when you long divide,you get a quotient that isn’t a fraction, plus a remainder that is afraction. The quotient isn’t a fraction – in polynomial long division,that means there’s a numerator but no denominator : look at theexample, where g(x) = x − 1. No part of that polynomial can go tozero.

Now the remainder: that’s a fraction. Being a fraction means that thenumerator is less than the denominator. Again, in polynomial terms,that means the degree of the numerator is less than the degree of thedenominator. Look at the example, where r(x) = 1/(x2 + x + 1); thelimit will be zero.

There is another way to think of f (x) = g(x) + r(x): it writes f as aglobal part plus a local part. :)

22 Kathy Davis

Figure 22: A Square Root Asymptote

To show you what the ’evil’ problemmight look like.

Level II Alert Here’s a different kind of problem: show f (x) =√x2 + x is asymptotic to g(x) = x + 1

2 . See Figure ??. Also, there willbe cuteness.

I need to compute limx→∞ [ f (x)− g(x)] = 0. Here goes:[√x2 + x− (x +

12)

]=

[√x2 + x− (x +

12)

]·[√

x2 + x + (x + 12 )√

x2 + x + (x + 12 )

]

Difference of Squares! Mine!

=(√

x2 + x)2 − (x + 12 )

2

1· 1√

x2 + x + (x + 12 )

=(x2 + x)− (x2 + x + 1

4 )

1· 1√

x2 + x + (x + 12 )

= −14· 1√

x2 + x + (x + 12 )

now the limit is easy.

Side Remark And how did I guess that g(x) = x + 12 ?? It’s Newton’s

Binomial Theorem, again: when a is small,

√1 + a = 1 +

12

a +12(

12− 1)

a2

2!+ · · · ≈ 1 +

12

a

The problem is, x is not small, we have x → ∞. Remember, though,we have a trick for that: we interchange zero and infinity by writingin powers of 1/x instead of powers of x. If x is large, then 1/x will besmall.


Section 1.9: Asymptotes and Removables: Practice

What we’re going to see is that the line between removable disconti-nuites and asymptotes can be difficult to find. A little more precisely:When we have a 1

0 we expect there to be an asymptotes, and whenwe have a 0

0 we expect there to be a removable discontinuity. Whatwe’re going to see is that this simple idea is too simple.

Figure 23: The Graph

Notice there is an asymptote whenx = 0, but a hole when x = 4..

Example

limx→4

(1x− 1

4

)(1− x + 1

x− 4

)When you plug in x = 4, you get

(14 −

14

) (1− 4+1

4−4

)and that 4+1

4−4

looks very much like the Asymptote Theorem asks: 50 . But, the

Asymptote Theorem does ask us to have a single quotient, numer-ator and denominator. This means we’ll have to bring 1

x −14 and

1− x+1x−4 each under a common denominator; then we’ll have to com-

bine the two. Here goes:

1x− 1

4=

4− x4x

; 1− x + 1x− 4

=(x− 4)− (x + 1)

x− 4=−5

x− 4

All together now:(1x− 1

4

)(1− x + 1

x− 4

)=

(4− x

4x

)(−5

x− 4

)=

5(x− 4)4x(x− 4)

So now when I plug in x = 4, I get 5(4−4)4·4(4−4) = 0

0 . This is a warningthat I might have a removable! Removable says I should cancel thex − 4 from numerator and denominator (I get 5

4x ), and then I shoulddo the limit:

limx→4

54x

=5

4 · 4 =5

16

There’s no asymptote at x = 4; there’s a removable discontinuity, asin Figure ??.

It’s worth thinking what went wrong here. When I first plugged inx = 4, I focused on the 4+1

4−4 part. But that’s just what the Asymptote

Theorem won’t let us do: Asymptote Theorem tells me I have to lookat the function as a whole, not one little piece at a time. And when Ilook at the whole, I see the

(14 −

14

)factor out in front: that’s a zero.

If I combine that zero with the 4+14−4 part, I get a 0 · 5

0 = 00 . Poster child

for removable.

24 Kathy Davis

Here’s another sneaky one:

Example

limx→−2

x2 − 4x2 + 4x + 4

Plug in to get 4−44+4·(−2)+4 = 0

0 , which, we’d say, is removable.

Removable theory tells me to factor out a common x − c; c = −2, sothat’s x− (−2) = x + 2 from numerator and denominator, cancel, andtake the limit again. Like this:

x2 − 4x2 + 4x + 4

=(x + 2)(x− 2)

(x + 2)2 =x− 2x + 2

limx→−2

x− 2x + 2

=−2− 2−2 + 2

=−40

Now the Asymptote Theorem does apply, and it tells me the functionhas a vertical asymptote at x = −2.

Functions that look like asymptotes but are removables. Functionsthat look like removables but are asymptotes. What’s next? Functionsthat look like jumps but are asymptotes?

How’s a gal supposed to tell the difference? By taking the limit:that’s your one true guide.

Oh, right: I should do this, then:

limx→3

x2 − 9|x2 − 6x + 9|

Or maybe you want to?

Calculus Sin Frontera · 2020-06-27 · calculus sin frontera 1 Section 1.1: Removable Discontinuities This mini-section is about doing limits for removable discontinuities. we actually

Documents