Page 1Page 1Why [0/0] and [ – ] Are Indeterminate Forms Whereas k/0 Isn’t?

[0/0] is an indeterminate form because a limit of this form is unpredictable. If you go back to the 8 problems on last two pages, each one is [0/0] and each one has a different answer. That means, it can be anything from a constant (zero or non-zero) to infinity (positive or negative) and it can be DNE also. Whereas, the limit of the form k/0 (k 0) is very predictable since it only can be ___________ (and can’t be anything else). Therefore, the (true) meaning of a limit in an indeterminate form is: A limit that can’t be done by direction substitution nor can it be predicted using direction substitution.

Now let’s look at why [ – ] is also an indeterminate form. Of course, one might wonder: Is it just 0? So let’s discuss it from Set Theory point of view.

What is the difference between the following sets? How is it related to limits?

1. {1, 2, 3, 4, 5, …} – {1, 2, 3, …} =

2. {1, 2, 3, 4, 5, …} – {4, 5, 6, …} =

3. {1, 2, 3, 4, 5, …} – {2, 4, 6, …} =

As you can see from the above, [ – ] is also as unpredictable as [0/0]. We will present you some examples in limits:

1

1

1lim.2

11lim.1

2120 xx

x

xx xx

Page 2More on the Indeterminate Form of [∞ – ∞]

Note: Most limit problems of the form [ – ] really come from [k/0 – k/0] (recall k/0 is either + or – on page 10). On the other hand, [k/0 – k/0] does not necessarily yield [ – ]. Let’s consider the following (where +k denotes a positive nonzero constant and –k denotes a negative nonzero constant):

[k/0 – k/0] [ – ]? Comment

1

1 13.lim

1( 1)x xx x

2 2 20 0 0

2 2 2 21 1 1 1

1 1 1 1 1 11. lim lim lim

0

1 1 1 ( 1) 1 12. lim lim lim lim

1 1 1 01 1 1 1

x x x

x x x x

x x

x x xx x x

x x x x x

x x xx x x x

20

1 14. lim

x x x

1. ( ) ( )0 0

2. ( ) ( )0 0

3. ( ) ( )0 0

4. ( ) ( )0 0

k k

k k

k k

k k

Page 3Calculating Limits Using the Limit Laws

Limit Laws Suppose that c is a constant and the limits lim f(x) and lim g(x) exist. Then

1.

2.

3.

4.

5. provided that lim g(x) 0

lim[ ( ) ( )] lim ( ) lim ( )x a x a x a

f x g x f x g x

lim[ ( ) ( )] lim ( ) lim ( )x a x a x a

f x g x f x g x

lim[ ( ) ( )] lim ( ) lim ( )x a x a x a

f x g x f x g x

lim[ ( ) / ( )] lim ( ) / lim ( )x a x a x a

f x g x f x g x

lim[ ( )] lim ( )x a x a

c f x c f x

Applications:Given: limx1 f(x) = 2, limx1 g(x) = –3, limx2 f(x) = 4 and limx2 g(x) = 0.Evaluate the following:

1. limx1 [f(x) + g(x)] 2. limx1 [2f(x) 3g(x)]

3. limx2 [f(x)g(x)] 4. limx2 [f(x)/g(x)] 5. limx2 [g(x)/f(x)]

xa xa

xa

Note: These problems are conceptual and intangible since neither the graphs nor the equations of the functions are given. Nevertheless, they are easy. Even without any of these limit laws, when you are given lim x1 f(x) = 2 and limx1 f(x) = –3, of course you are going to evaluate limx1 [f(x) + g(x)] as 2 + (3) = 1 (what else can it be?)

Sum Law

Difference Law

Product Law

Quotient Law

Constant Multiple Law

Page 4Calculating Limits Using the Limit Laws (cont’d)

Limit Laws Suppose that the limits lim f(x) and lim g(x) exist. Then

6.

7. if n is even, lim f(x) 0

Applications:I. Given: limx1 f(x) = 2, limx2 f(x) = 8 and limx3 f(x) = 4. Evaluate the following:

1. limx1 [f(x)]4 2. limx2 3. limx3

xa xa

Root Law

Power Lawn

ax

n

axxfxf )](lim[)]([lim

nax

n

axxfxf )(lim)(lim

xa

3 )(xf )(xf

II. Evaluate: limx3 (2x + 4)5

y

xO

Use the graph and limit laws to evaluate the following limits:

f

g

1. limx1 [f(x) + g(x)] 2. limx3 [2f(x) 3g(x)]

3. limx5 [f(x)g(x)] 4. limx3 [f(x)/g(x)]

5. limx4 [f(x) + g(x)]

Page 5The Squeeze Theorem

The Squeeze Theorem If f(x) g(x) h(x) when x is near a (except possibly at a) and

then

Lxhxfaxax

)(lim)(lim

lim ( )x a

g x L

In layman’s terms, the Squeeze Theorem says if g(x) is squeezed in between f(x) and h(x) near a, and if f and h have the same limit L at a, then g is forced to have the same limit at a also (see Figure 9). The Squeeze Theorem, a.k.a. the Sandwich Theorem or the Pinching Principle, allows us to evaluate limits that are not in any previous prescriptions:

Example: Use the Squeeze Theorem to show that . 01

sinlim 2

0

xx

x

0 a

f

g

hL

Figure 9

Note: Notice that limx0 sin (1/x) = sin () = DNE (see graph of y = sin x below), therefore, limx0 x2 sin (1/x) = limx0 x2limx0 sin (1/x) would be 0DNE, which is a special format not mentioned previously.

y = sin x

y = x2

y = –x2

y = x2 sin (1/x)

1

–1

Page 6Limits Involving Absolute Value

We first saw a problem involving absolute value on page 12 (problem 3 to be exact). Since every problem on that page is evaluated using the tabular method. The question is: How do we evaluate limit problems involving absolute value without using the tabular method? The answer is: Algebraically, with a split consideration.

Recall the basic algebraic definition of |x|:

The fact is: where c is

called the cutoff number. To determine this cutoff number, we set the expression = 0 and solve for x.

Example: since 2x + 5 = 0 x =

0,

0,||

xx

xxx

e pression,e pression

(e pression),

x x cx

x x c

,2 5

,

xx

x

Q: How does this relate to limit problems involving absolute values?A: When the function involves absolute value, try to rewrite it without absolute value.

Examples:

3

31.lim

3x

x

x

22

22.lim

3 2x

x

x x

2

1

5 43. lim

1x

x x

x

(See

abo

ve)

Note: Not all limit problems involving absolute values require the split consideration. For example, we do not need to (and we don’t) split the absolute-value expression into two non-absolute-value expressions when direct substitution is sufficient or when it yields k/0 (on the other hand, we do need to split it when direct substitution yields 0/0).Examples: 1. limx–2 |2x + 5| = 2. limx–2 3/|2x + 4| =