Calculus Lectures: Determinate Forms

by Kathy Davis

Copyright ©2021 Kathy Davis

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Written February 2021

2 Kathy Davis

Figure 1: GPS Satellites

To know where I am, on the earth, thesatellite needs to know where is, tovery high accuracy.

Now for something completely different: I’m heading towards asection where I look at computing functions to any desired accuracy.For example, GPS positioning, Figure 1, needs over fifty digits forsome calculations. To do these kinds of computations, I’ll be doing alot of limits. So, this lecture is a review of limits.

The precise definition of a limit is this:

limx→c

f (x) = L means, for every ε > 0 you can find a δ > 0 such that

for all x with 0 < |x− c| < δ it must be that | f (x)− L| < ε

The phrase "for all" is important: it’s not enough for a few x to getf (x) close to L. I get close to L and I stay close to L.

Now I’ll need two more limit definitions, also from first semestercalculus:

limx→c

f (x) = ∞ and limx→∞

f (x) = L

The intuitive meaning of the first is that when x and c are close, thenf (x) gets large and stays large. How large? That’s just like asking’how close’ for ordinary limits. I get told how large it needs to be,and I reply what δ is needed to make it that large:

limx→c

f (x) = ∞ means, for every M > 0 I can find a δ > 0 such that

for all x with 0 < |x− c| < δ it must be that f (x) > M

M tells me how large f has to be, and δ tells me what I need to makethat happen.

For the second, as x gets larger and larger, then f (x) and L get close.Again, ε tells me how close, and B will tell me how large x needs tobe:

limx→∞

f (x) = L means, for every ε > 0 I can find a B such that

for all x > B it must be that | f (x)− L| < ε

To see how these work, try

limx→0+

1x= ∞

So, let’s start with a B > 0, and I have to find a δ and I have to show0 < x < δ makes 1

x bigger than B (I replaced 0 < |x − 0+| < δ with0 < x < δ because the limit is to 0+ so 0 < x). Anyway: I chooseδ = 1

B . Then 0 < x < δ gives 0 < x < 1B , flipping, 1

x > B which iswhat I’m supposed to show.

calculus lectures 3

This is a bit – well, tedious; what I want to do in the lecture is changethese kinds of computations into something more like algebra. Also,more like: faster.

So I showed in detail that

limx→0+

1x= ∞

In fact, a lot more is true:

If limx→ c

f (x) = 0+, Then limx→ c

1f (x)

= ∞

This isn’t a limit; it’s a whole collection of limits. It’s called a formbecause I can put all kinds of things into the ’ f ’ slot: this limit "form"wants me to fill in a function, and if lim f (x) = 0+, I’ll get positiveinfinity.

These kinds of results are called determinate forms, and there’s a short-hand for them:[

If limx→ c

f (x) = 0+, Then limx→ c

1f (x)

= ∞]↔[

10+

= ∞]

The statement 10+ = ∞ looks like algebra, like saying 1

2 = .5. It isNOT algebra. It is a shorthand notation for an infinite collection oflimits, a fast way of writing determinate forms.

These determinate forms are going to save everyone a lot of time:that’s what shorthands do! For example, I’ll write determinate formslike ∞ + ∞ = ∞, which is shorthand for

If limx→ c

f (x) = ∞ and limx→ c

g(x) = ∞, Then limx→ c

[ f (x) + g(x)] = ∞

There are also indeterminate forms; one example is∞−∞ = indeterminate. What this means that

If limx→ c

f (x) = ∞ and limx→ c

g(x) = ∞, Then limx→ c

[ f (x)− g(x)] =

Equals what? What it equals depends on what f and g are, and ifyou choose different f and g, you can get different answers:

limx→ ∞

[x− x] = limx→ ∞

0 = 0

limx→ ∞

[2x− x] = limx→ ∞

x = ∞

limx→ ∞

[x− 2x] = limx→ ∞

−x = −∞

So: lots of different answers means ∞−∞ = indeterminate.

4 Kathy Davis

This lecture works with the determinate forms; the next two lectureswill do the indeterminate forms.

Here’s a list of the determinate forms I need to do my work. Thereare more, but these are enough. First, some notation: the symbol α inmany of them stands for a finite, fixed (no x in it) real number, like2 or something. Also, α cannot be ±∞. "dne" means "does not exist".Oh also: "detform" means "determinate form".

∞± α = ∞

α ·∞ =

+∞ if α > 0−∞ if α < 0indeterminate if α = 0

∞ ·∞ = ∞

(−∞) ·∞ = −∞

1+∞

= 0+1−∞

= 0−

10+

= ∞1

0−= −∞ so

10

dne

∞α =

∞ if α > 00 if α < 0indeterminate if α = 0

e∞ = ∞ e−∞ = 0+

ln(0+) = −∞ ln(∞) = ∞ ln(0) dne

∞−∞ = indeterminate∞∞

= indeterminate

00= indeterminate

0 ·∞ = indeterminate

The point of having forms is to make limits easy; here’s an example:

limx→∞

e−x2= e−∞2

= e−∞ since ∞2 = ∞; then e−∞ = 0+ is a detform

I’d just writelim

x→∞e−x2

= e−∞ = 0+

calculus lectures 5

limx→∞

(x2 + 2) = ∞2 + 2 = ∞ + 2 = ∞

I’d just writelim

x→∞(x2 + 2) = ∞

But, I have to be careful:

limx→∞

(x2 − 3x) = ∞2 − 3 ·∞−∞ = indeterminate

Rewrite limx→∞

(x2 − 3x) = limx→∞

x(x− 3) = ∞(∞− 3) = ∞ ·∞ = ∞

I’d just writelim

x→∞(x2 − 3x) = lim

x→∞x(x− 3) = ∞

WARNING Don’t make up determinate forms. For example, some-times I see 0∞ = 0 or 2∞ = ∞. Then I ask where did those come fromand I get "well zero to any power is zero" or "2 to bigger powers getsbigger and bigger".

What, this is an essay course now?

So look at limx→0 x1x , say for x = − 1

10 ( and so 1x = −10):

[− 1

10

]−10= [−1]−10

([110

]−1)10

= 1010

That’s – ten billion. Ten billion is so zero – right, sure it is. You beensmoking weird stuff? You can’t make up your own detforms. Put thelist of detforms on a cheat sheet and stick to it.

Worked Problems

limx→∞

(x2 + 1)7

x14 =∞∞

= indeterminate

I need to cancel x’s from numerator and denominator, until there’sno more indeterminate ∞

∞[x2 + 1

]7x14 =

[x2 + 1

]7[x2]

7 =

[x2 + 1

x2

]7

=

[1 +

1x2

]7

limx→∞

(x2 + 1)7

x14 = limx→∞

[1 +

1x2

]7=

[1 +

1∞2

]7= [1 + 0]7 = 1

I’d just write

limx→∞

(x2 + 1)7

x14 = limx→∞

[1 +

1x2

]7= [1]7 = 1

WARNING: Here’s what I cannot do:

limx→∞

(x2 + 1)7

x14 ≈ x14

x14 → 1

6 Kathy Davis

Why not?

First: When I changed limx→∞(x2+1)7

x14 to x14

x14 , where’d the limit go?Leaving "limit" out: -3 points each time.

Second: What does→ mean? If it mean equals, write equals. Again,-2 points for each missing = sign.

Third: What does ≈ mean? Loss of all points for using this. Why?

limx→∞

(x + 1)x

xx ≈ xx

xx → 1

It’s the same thing, right? But actually,

limx→∞

(x + 1)x

xx = e

So this ≈ thing sometimes gives right answers and sometimes wronganswers. Do I know which? Do I know why? That is to say: if I don’tunderstand what I’m doing, or why, or whether answers are right orwrong – what good am I? Can I go home now?

WARNING

limx→∞

(x2 + 1)7

x14 = limx→∞

x14

x14 means loss of all points. Like a car crash : Totalled

Worked Problems

limx→0+

ln xx

=ln 0+

0+=−∞0+

= [−∞]

[1

0+

]= [−∞] [∞] = −∞

What I did was manipulate the −∞0+ algebraically until I got detforms

to finish the problem.

Problem limx→∞

x√x2 + 1

=∞∞

and once again I have to cancel the infinities, as I did in the firstproblem. One way to cancel would give square roots tonumeratorand denominator. So, since x > 0, x =

√x2, and

x√x2 + 1

=

√x2

√x2 + 1

=

√x2

x2 + 1=

√√√√ x2

x2

x2+1x2

=

√1

1 + 1x2

limx→∞

x√x2 + 1

= limx→∞

√1

1 + 1x2

=

√1

1 + 1∞2

=

√1

1 + 0= 1

Or, factor out common powers of x to cancel:

x√x2 + 1

=x√

x2(

1 + 1x2

) =x

√x2

√(1 + 1

x2

) =x

x√(

1 + 1x2

) =1√(

1 + 1x2

)Now take the lmit just like the first way. Also, I chose these first threeproblems so that using L’Hospital would be painful or impossible (I’llshow that in the next lecture).

calculus lectures 7

Problem limx→∞

x2 − 2x + 3x3 − x− 1

If I rewrite the numerator as x(x − 2) + 3 and the denominator asx(x2 − 1)− 1, I’d see the limit of each of these is infinity, so now I canuse L’Hospital.

WARNING I must rewrite, so L’Hop isn’t the fastest way on this.

I’ll skip L’Hop and divide numerator and denominator by x3.

x2 − 2x + 3x3 − x− 1

=1x −

2x2 +

3x3

1− 1x2 − 1

x3

limx→∞

x2 − 2x + 3x3 − x− 1

=0− 0 + 01− 0− 0

=01= 0

If I divide numerator and denominator by x2:

x2 − 2x + 3x3 − x− 1

=1− 2

x + 3x2

x− 1x −

1x2

limx→∞

x2 − 2x + 3x3 − x− 1

=1− 0 + 0∞− 0− 0

=1∞

= 0

How does a computer do these limits?

Laurent Series? Are you kidding? Not so much – I’ll see later in thecourse what a Laurent series is, and how to use it to compute a limit.

8 Kathy Davis

Until then – time for a bit of practice, at Kathy’s 14U Show Place ofFun.