Finding Limits Graphically & Numerically

Chapter 2.2

An Introduction to Limits

• • Consider the function 𝑥 3 − 1 𝑓 𝑥 = 𝑥 − 1 The graph of the curve is a parabola, but it has an unusual feature: a “hole” at 𝑥 = 1 . Why?

• We can evaluate

f

for any value other than 𝑥 = 1 • What happens at 𝑥 = 1 ?

• More specifically, as the values of

x

get closer to 1 from both the left and the right, what happens to the function values?

An Introduction to Limits

Evaluate the function 𝑓 𝑥 = 𝑥 3 −1 𝑥−1 for the following values of

x

:

x

f

(

x

)

0.75

0.9

0.99

0.999

1 ?

1.001

1.01

1.1

1.25

An Introduction to Limits

Evaluate the function 𝑓 𝑥 = 𝑥 3 −1 𝑥−1 for the following values of

x

:

x

0.75

0.9

0.99

0.999

f

(

x

)

2.313

2.710

2.970

2.997

1 ?

1.001

1.01

1.1

1.25

3.003

3.030

3.310

3.813

An Introduction to Limits

• Although the function is not defined at 𝑥 = 1 , we can study the behavior of the function as the values of

x

approach 1 • • • We use the following notation (you must know this!) lim 𝑥→1 𝑓 𝑥 = 3 This is read as “the limit as

x

approaches 1 of lim 𝑥→𝑐 𝑓(𝑥) = 𝐿 𝑓(𝑥) is equal to 3” In general, if a function 𝑓(𝑥) becomes arbitrarily close to some number

L

as say that the

x

approaches some number

c

limit

of 𝑓(𝑥) as

x

approaches

c

from either side, then we is

L

• The notation for this is

Example 1: Estimating a Limit Numerically

Evaluate the function 𝑓 𝑥 = 𝑥 at several points near 𝑥 = 0 𝑥+1−1 use the results to estimate the limit lim 𝑥→0 𝑥 𝑥+1−1 .

Create a table with values close to zero on either side (i.e., positive and negative).

and

x

−𝟎. 𝟎𝟏 −𝟎. 𝟎𝟎𝟏 −𝟎. 𝟎𝟎𝟎𝟏

0

𝟎. 𝟎𝟎𝟎𝟏 𝟎. 𝟎𝟎𝟏 𝟎. 𝟎𝟏 𝑓(𝑥)

Example 1: Estimating a Limit Numerically

Evaluate the function 𝑓 𝑥 = 𝑥 at several points near 𝑥 = 0 𝑥+1−1 use the results to estimate the limit lim 𝑥→0 𝑥 𝑥+1−1 .

Create a table with values close to zero on either side (i.e., positive and negative).

and

x

−𝟎. 𝟎𝟏 −𝟎. 𝟎𝟎𝟏 −𝟎. 𝟎𝟎𝟎𝟏

0

𝟎. 𝟎𝟎𝟎𝟏 𝟎. 𝟎𝟎𝟏 𝟎. 𝟎𝟏 𝑓(𝑥) You can see that, as

x

approaches zero, the function values approach 2. So we write as an answer 𝑥 lim 𝑥→0 𝑥 + 1 − 1 = 2 Note that a limit does not depend on whether or not a function is defined at a given point!

Example 2: Finding a Limit

Find the limit of 𝑓(𝑥) as

x

approaches 2, where

f

𝑓 𝑥 = 1, 𝑥 ≠ 2 0, 𝑥 = 2 is defined as

Example 2: Finding a Limit

Find the limit of 𝑓(𝑥) as

x

approaches 2, where

f

𝑓 𝑥 = 1, 𝑥 ≠ 2 0, 𝑥 = 2 is defined as As the graph shows, if

x

can write approaches 2 from either the left or the right, the function value approaches 1. So we lim 𝑥→2 𝑓(𝑥) = 1 Note that this is true

even though

𝑓 2 = 0

by definition

. To reemphasize, when finding the limit of a function, it does not matter whether the function is or is not defined at the given point.

An Introduction to Limits

• In the previous examples you used a numerical (or tabular) approach and a graphical approach to finding a limit • These are often helpful and it is recommended that you explore one or both of these methods when solving problems • The most important method, however, is the analytic approach; using either algebra or calculus or both • Only the analytic approach allows us to conclude with certainty • Whenever we speak of “doing calculus”, we are talking mainly about the analytic approach

Limits That Fail to Exist

• Function values, when they exist are always unique; for any given function, as values approach some value of

x

, the limit of the function must approach one and only one value • However, limits may fail to exist • The next three example show three of the most common ways that a limit may fail to exist

Example 3: Behavior That Differs from the Right & Left

For the function 𝑓 𝑥 = 𝑥 𝑥 , show that lim 𝑥→0 𝑓(𝑥) does not exist (DNE).

Example 3: Behavior That Differs from the Right & Left

For the function 𝑓 𝑥 = 𝑥 𝑥 , show that lim 𝑥→0 𝑓(𝑥) does not exist (DNE).

Note first that the function is not defined at 𝑥 = 0 . But as

x

not exist (DNE).

approaches zero from the positive (i.e., the right) side, the function approaches 1. As

x

approaches zero from the negative, or left, side, the function approaches negative one. Since the value of a function must be unique, we say that the function as

x

approaches zero does We will use the following notation in section 2.4

𝑥 lim 𝑥→0 + 𝑥 = 1, lim 𝑥→0 − 𝑥 𝑥 = −1 The first equation is read, “the limit as

x

approaches zero from the positive side”. That is, as from values 𝑥 > 0 . The second equation is read, “the limit as

x x

approaches zero approaches zero from the negative side”. That is, as

x

approaches zero from values 𝑥 < 0 . Since the

one-sided

limits do not agree, the limit does not exist.

Example 4: Unbounded Behavior

Discuss the existence of the limit 1 lim 𝑥→0 𝑥 2

Example 4: Unbounded Behavior

Example 4: Unbounded Behavior

Example 4: Unbounded Behavior

Discuss the existence of the limit 1 lim 𝑥→0 𝑥 2 As the values of

x

approach zero, the function values increase without bound. We will later be able to write 1 lim 𝑥→0 𝑥 2 = ∞ This is the same as saying that the limit DNE. However, the above equation is not really justified at this point. Infinity is not a number, so we may not assume that it behaves like a real number and we should not carelessly apply symbols that are reserved for numbers. Among the many things you will learn in this course, communicating your ideas clearly and accurately is one of the most important!

In a later chapter we will specifically define what it means to say something equals infinity. Then the above equation will be justified.

Example 5: Oscillating Behavior

Discuss the existence of the limit lim 𝑥→0 sin 1 𝑥

Example 5: Oscillating Behavior

Discuss the existence of the limit lim 𝑥→0 sin 1 𝑥

x

sin 1 𝑥 𝟐 𝝅 ≈ 𝟎. 𝟔𝟑𝟕 𝟐 𝟑𝝅 ≈ 𝟎. 𝟐𝟏𝟐 𝟐 𝟓𝝅 ≈ 𝟎. 𝟏𝟐𝟕 𝟐 𝟕𝝅 ≈ 𝟎. 𝟎𝟗𝟏 𝟐 𝟗𝝅 ≈ 𝟎. 𝟎𝟕𝟏 𝟐 𝟏𝟏𝝅 ≈ 𝟎. 𝟎𝟓𝟖 1 −1 1 −1 1 −1

Example 5: Oscillating Behavior

Discuss the existence of the limit lim 𝑥→0 sin 1 𝑥 As

x

approaches zero, the function values oscillate between 1 and −1 . The never settle on one particular value. Hence, the limit DNE. This type of behavior is most often seen in trigonometric functions.

Limits That Fail to Exist

We will encounter these kinds of limits that fail to exist, so remember the three types (where

c

is the value that

x

approaches) 1.

2.

3.

𝑓(𝑥) approaches a different number from the right side of

c

approaches from the left side than it 𝑓(𝑥) increases or decreases without bound as

x

approaches

c

𝑓(𝑥) oscillates between two fixed values as

x

approaches

c

A Formal Definition of Limits

• • • • As we have seen, for a function 𝑓(𝑥) , if, as 𝑥 approaches some value

c

, the function approaches some value

L

, then our notation is lim 𝑥→𝑐 𝑓(𝑥) = 𝐿 However, we must give a mathematical interpretation to the phrases “

x

approaches

c

” and “ 𝑓(𝑥) approaches

L

” To do this, we will choose some arbitrary positive number 𝜀

epsilon

) and require that the function values fall within 𝜀 of (Greek

L

. Symbolically, 𝑓 𝑥 − 𝐿 < 𝜀 or 𝐿 − 𝜀 < 𝑓 𝑥 < 𝐿 + 𝜀 Note that when 𝜀 is very small, we will have a “target” of values close to

L

; the target becomes narrower as 𝜀 becomes smaller

A Formal Definition of Limits

• The phrase “

x

number 𝛿 approaches (Greek

delta

) of

c c

” will mean that

x

is within a positive • • • In addition, we will

not

require that 𝑥 = 𝑐 the function is not defined at 𝑥 = 𝑐 ) • Our notation is 0 < 𝑥 − 𝑐 < 𝛿 or 𝑐 + 𝛿 < 𝑥 < c − 𝛿, 𝑥 ≠ 𝑐 Note that 0 < |𝑥 − 𝑐| be defined at 𝑥 = 𝑐 (to avoid the possibility that means that we do not require that the function As previously, the smaller the value of 𝛿 , the narrower is the “target” of values near

c

A Formal Definition of Limits

A Formal Definition of Limits

A Formal Definition of Limits

DEFINITION: Let

f

be a function defined on an open interval containing

c

possibly at

c

) and let

L

be a real number. The statement lim 𝑥→𝑐 𝑓(𝑥) = 𝐿 means that for each 𝜀 > 0 there exists a 𝛿 > 0 such that, if (except 0 < 𝑥 − 𝑐 < 𝛿, then 𝑓 𝑥 − 𝐿 < 𝜀

A Formal Definition of Limits

• • • • In words, this means all of the following We are allowed to

choose

any positive value of 𝜀 Having chosen 𝜀 , we must be able to

find

some value of 𝛿 For any

x

in the interval function value 𝑓(𝑥) (𝑐 − 𝛿, 𝑐 + 𝛿) not equal to

c

, there must be a in the interval (𝐿 − 𝜀, 𝐿 + 𝜀) If this condition is met (i.e., finding 𝛿 ) for

any

justified in saying that the limit is the value

L

value of 𝜀 , we are Another way to think of this definition is this: for values of

x

there will always be values of 𝑓(𝑥) near

L

near

c

, no matter how small we , choose 𝜀 to be

A Formal Definition of Limits

A word about proper notation. Consider the function we saw earlier, 𝑓 𝑥 = 𝑥 3 − 1 𝑥 − 1 It is

never

correct to write 𝑓 1 = 3 . This is not a true statement because the function is not defined at 𝑥 = 1 . But it

is

lim 𝑥→1 𝑓(𝑥) = 3 correct to write because we have

defined

what it means for a limit to equal some value. We are not justified in using an equal sign in the first place, but we are fully justified in the second place.

Example 6: Find a 𝛿 for a Given 𝜀

Given the limit lim 𝑥→3 (2𝑥 − 5) = 1 , find a 𝛿 such that 2𝑥 − 5 − 1 < 0.01

whenever 0 < 𝑥 − 3 < 𝛿 .

2𝑥 − 5 − 1 < 0.01 ⟹ 2𝑥 − 6 < 0.01 ⟹ 2 𝑥 − 3 < 0.01 ⟹ 𝑥 − 3 < 0.005

Hence, if we use 𝛿 = 0.005

, then 0 < 𝑥 − 3 < 0.005

implies that 2𝑥 − 5 − 1 = 2 𝑥 − 3 < 2 0.005 = 0.01

which is what we are required to show. The value of 𝛿 value will also work.

here is the

largest

such value. Any smaller positive Note that this example does NOT prove that the limit is 1. To say that the limit exists and is equal to 1, we must be able to find 𝛿 for

any

𝜀 .

Example 7: Using the 𝜀−𝛿 Limit Definition of a

Use the 𝜀−𝛿 definition of a limit to prove that lim 𝑥→2 3𝑥 − 2 = 4 In order to use the definition, we must do two things. First, we must find a way to express 𝛿 of 𝜀 > 0 . Second, we must show that, for the given 𝛿 , 0 < 𝑥 − 2 < 𝛿 ⟹ for any given value 3𝑥 − 2 − 4 < 𝜀 .

To find 𝛿 (which will be a function of 𝜀 , though it need not be), we use the inequality simplifies as follows 3𝑥 − 2 − 4 < 𝜀 3𝑥 − 6 < ε 3 𝑥 − 2 < 𝜀 𝜀 𝑥 − 2 < 3 3𝑥 − 2 − 4 < 𝜀 . This Note that we have been able to simplify the inequality so that the left side looks exactly like (except for the 0 < ) 𝜀 the inequality 𝑥 − 2 < 𝛿 . So we can choose 𝛿 = . Next we must show that, using this value of 𝛿 , the 3 implication holds.

Example 7: Using the 𝜀−𝛿 Limit Definition of a

Use the 𝜀−𝛿 definition of a limit to prove that lim 𝑥→2 3𝑥 − 2 = 4 We say, then, that if 0 < 𝑥 − 2 < 𝛿 , then 𝜀 3𝑥 − 2 − 4 = 3 𝑥 − 2 < 3𝛿 = 3 ⋅ 3 = 𝜀 Therefore 3𝑥 − 2 − 4 < 𝜀 and we have proven that the limit is, indeed, 4.

Example 8: Using the 𝜀−𝛿 Limit Definition of a

Use the 𝜀−𝛿 definition of a limit to prove that lim 𝑥→2 𝑥 2 = 4 Again, we must do two things: find a value for 𝛿 , then show that for this value of 𝛿 0 < 𝑥 − 2 < 𝛿 ⟹ 𝑥 2 − 4 < 𝜀 Note from the previous example that we must connect |𝑥 − 2| 𝑥 2 − 4 < 𝜀 𝑥 + 2 𝑥 − 2 to 𝑥 + 2 𝑥 − 2 < 𝜀 |𝑥 < 𝜀 2 − 4| . We start by factoring Now we have part of what we want, but we also have another factor, containing

x

how to proceed. What we will do is find a reasonable estimate for what |𝑥 + 2| and it’s not immediately clear might be, given that we want values of

x

that are close to 4. That is, we want to find a number

K

so that 𝑥 + 2 < 𝐾 . Then our inequality becomes 𝜀 𝐾 𝑥 − 2 < 𝜀 ⟹ 𝑥 − 2 < 𝐾

Example 8: Using the 𝜀−𝛿 Limit Definition of a

Use the 𝜀−𝛿 definition of a limit to prove that lim 𝑥→2 𝑥 2 = 4 So now we have 𝜀 𝐾 𝑥 − 2 < 𝜀 ⟹ 𝑥 − 2 < 𝐾 To find a value for

K

, we remember that we need only be concerned with values of

x

assume the following 𝑥 − 2 < 1 that are close to 2. We will This merely says that he values of

x

are within 1 of the number 2, which is not unreasonable. But this seems like a cheat because you cannot assume what you set out to find. In fact, it may turn out that for some values of

x

within 1 of 2, the function values will

not

fall within 𝜀 of

L

. But we’ll use a clever way to avoid this possibility.

Now we must find an inequality involving |𝑥 + 2| that is implied by 𝑥 − 2 < 1 .

Example 8: Using the 𝜀−𝛿 Limit Definition of a

Use the 𝜀−𝛿 definition of a limit to prove that lim 𝑥→2 𝑥 2 = 4 Remove the absolute value to get −1 < 𝑥 − 2 < 1 ⟹ 1 < 𝑥 < 3 Now add 2 2 − 1 < 𝑥 + 2 < 3 + 2 ⟹ 1 < 𝑥 + 2 < 5 Note that since 0 < 1 < 𝑥 + 2 , then we can safely write have made two assumptions, both of which must be true 𝑥 + 2 < 5 = 𝐾 𝜀 𝑥 − 2 < 1 AND 𝑥 + 2 < 5 . At this point, stop to notice that we We must use both inequalities in choosing 𝛿 . Specifically, we will define delta as 𝜀 𝛿 = min 1, 5

Example 8: Using the 𝜀−𝛿 Limit Definition of a

Use the 𝜀−𝛿 definition of a limit to prove that lim 𝑥→2 𝑥 2 = 4 Then, if 1 < 𝜀 5 we must have 𝜀 we will have 𝑥 − 2 < 1 < as part of the inequality. 𝜀 5 ; if 𝜀 5 < 1 we will have 𝑥 − 2 < 𝜀 5 < 1 . The important point is that Now, if 0 < 𝑥 − 2 < 𝛿 , then 𝜀 𝑥 − 2 < 1 AND 𝑥 − 2 < 5 but 𝑥 − 2 < 1 ⟹ 𝑥 + 2 < 5 Finally, 𝑥 2 𝜀 − 4 = 𝑥 + 2 𝑥 − 2 < 5 𝑥 − 2 < 5 ⋅ 5 = 𝜀 And we are done.

Extra Example: Using the 𝜀−𝛿 a Limit Definition of

Use the 𝜀−𝛿 definition of a limit to prove that 2 lim 𝑥→3 𝑥 + 3 = 1 3 value for 𝜀 − = = = 2 1 = 𝑥 − 3 𝑥 + 3 3 3 𝑥 + 3 3 𝑥 + 3 3 𝑥 + 3 3 𝑥 + 3 < 𝜀 We must find an approximation for the denominator. To do this, we start by assuming that 𝑥 − 3 < 1 We can choose any number here, but 1 is convenient to work with. Make sure you understand

why

the end, we will pick a value of 𝛿 that is the smaller of two values, guaranteeing |𝑥 − 3| this is ok: in will be less than both values at the same time.

Extra Example: Using the 𝜀−𝛿 a Limit Definition of

Use the 𝜀−𝛿 definition of a limit to prove that 2 lim 𝑥→3 𝑥 + 3 = 1 3 Next, we follow up on the implication of taking 𝑥 − 3 < 1 𝑥 − 3 < 1 ⟹ −1 < 𝑥 − 3 < 1 ⟹ 2 < 𝑥 < 4 ⟹ 5 < 𝑥 + 3 < 7 Because the approximation is in the denominator, we need to take a reciprocal. If two number

a

1 same sign, then 𝑎 < 𝑏 ⟹ 𝑏 < 1 𝑎 . So now we have 1 7 < 1 𝑥 + 3 < 1 5 ⟹ 1 𝑥 + 3 < 1 5 and

b

are the Now we go back to where we stopped before and write 𝑥 − 3 < 𝜀 ⟹ 𝑥 − 3 < 15𝜀 3 ⋅ 5

Extra Example: Using the 𝜀−𝛿 a Limit Definition of

Use the 𝜀−𝛿 definition of a limit to prove that 2 lim 𝑥→3 𝑥 + 3 = 1 3 Now we choose 𝛿 = min{1,15𝜀} . Finally, we show that the implication required by the definition holds If 0 < 𝑥 − 3 < 𝛿, then 𝑥 − 3 < 1 AND 𝑥 − 3 < 15𝜀 But the right half of this conjunction has further implications 1 𝑥 − 3 < 1 ⟹ 𝑥 + 3 To conclude, 2 𝑥 + 3 − 1 3 𝑥 − 3 = 3 𝑥 + 3 < 1 < 𝑥 − 3 ⋅ 3 𝑥 + 3 1 5 < 1 3 ⋅ 5 ⋅ 15𝜀 = 𝜀 And we are done.

Exercise 2.2a

• Page 74, #1-30

Exercise 2.2b

• Page 76, #31-38, 35-45 odds, 51-58, 59-61