Epsilon-Delta

YOU WILL LEARN ABOUT:

How to use the epsilon-delta definition of the limit to prove the limit of a function $f$ as $x$ approaches $a$ is $L$ .

The precise meaning of

$\lim_{x \to a} f (x) = L$ states that for every number

$ε > 0$ , there is a number

$δ > 0$ such that

$0 < |x - a| < δ$ then

$|f (x) - L| < ε$ .

The limits are important in many applications. The definition is particularly useful in situations where the margin of error is of particular importance, such as in the following examples.

The landing speed of the Mars Rover is allowed to vary by

m/s. Using the

definition of limits, a NASA engineer can determine the maximum number of meters,

, above the Martian surface the rocket initiation can deviate from the ideal height in order to be within our tolerable margin of error.

A nurse must give a certain amount of medication to a patient based on the patient's weight. If the margin of error in dosage is

mg, then the nurse can use the

definition of the limit to figure out the maximum number of ounces,

, her weight measurement of the patient can be off to safely and effectively medicate the patient.

The internal temperature of a Thanksgiving turkey should be within

F of some ideal temperature. Using the

definition of the limit, a chef can determine the maximum time,

, the cooking time can differ from its ideal value to keep the turkey's internal temperature within its allowable range of values.

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Note

Note that the disk method is not applicable here since not all the slices of the solid generated are disks. However, you will be able to compute its volume using washer method which will be covered in the next module.

THE ε-δ DEFINITION OF A LIMIT

, however,

. Increase ε or decrease δ.

δ₂

SHOW EPSILON & DELTA RECTANGLES

Yes No

Function 1:

$Function 3 : f (x) = 2 x - 1$

Function 2:

$Function 2 : f (x) = \sqrt{x}$

Function 3:

$Function 3 : f (x) = \frac{x^{3} - 2 x^{2}}{x - 2}$

Function 4:

$Function 4 : f (x) = \{\begin{matrix} x + 1 if x < 1 \\ x - 1 if x \geq 1 \end{matrix}$

$a =$

$L = 3$

$\lim_{x \to a} f (x) = L means if 0 < |x - a| < δ, then |f (x) - L| < ε .$

$δ = \min (δ_{1,} δ_{2}) (since f (x) is nonlinear)$

$δ = \min (δ_{1,} δ_{2}) (since f (x) has a jump)$

1 2 3

1. According to the ε - δ definition of the limit, if

and ε = 1, there is a value of δ for which |x - 2| < δ implies
|(2x - 1) - 3| < ε. What is this value of δ?

δ = 2

δ = 1/2

δ = 4

δ = 1/4

No such δ exists

2. According to the ε - δ definition of the limit, if

and ε = 0.5, there is a value of δ for which |x - 0| < δ implies
|(2x - 1) - (-1)| < ε. For what value(s) of x does this hold?

x = 0

-0.5 < x <0.5

-0.25 < x <0.25

-1 < x <1

No such δ exists

3. According to the ε - δ definition of the limit, if

and ε = 1, there is a value of δ for which |x - (-2)| < δ implies
|(2x - 1) - (-3)|<ε. What is this value of δ?

δ = 4

δ = 1/4

δ = 2

δ = 1/2

No such δ exists

1 2 3

1. According to the ε - δ definition of the limit, if

and ε = 0.5, there is a value of δ for which 0 < |x - 4| < δ implies

. What is the maximum value of δ₁ for which this is true?

δ₁ = 1.75

δ₁ = 2

δ₁ = 2.25

δ₁ = 4

No such δ exists

2. According to the ε - δ definition of the limit, if

and ε = 0.5, there is a value of δ for which |x - 4| < δ implies

. What is the maximum value of δ₂ for which this is true?

δ₂ = 1.75

δ₂ = 2

δ₂ = 2.25

δ₂ = 1.25

No such δ exists

3. Based on your answers to questions 1 and 2, what value should you choose for the value of δ to prove that

δ = δ₁

δ = δ₂

δ = δ₁ + δ₂

δ = δ₁ - δ₂

No such δ exists

1 2 3

1. According to the ε - δ definition of the limit, if

and ε = 1, there is a value of δ for which
|x - 2| < δ implies |f(x) - 4| < ε. What is the maximum value of δ?

δ ≈ 0.236

δ ≈ 0.267

δ ≈ 0.577

δ ≈ 0.447

No such δ exists

2. According to the ε - δ definition of the limit, if

and ε = 0.5, there is a value of δ for which |x - 2| < δ implies |f(x) - 4| < ε. What is the maximum value of δ?

δ ≈ 0.129

δ ≈ 0.121

δ ≈ 0.471

δ ≈ 0.535

No such δ exists

3. According to the ε - δ definition of the limit, if

and ε = 0.5, there is a value of δ for which |x - (-2)| < δ implies |f(x) - 4| < ε. What is the maximum value of δ?

δ ≈ 0.471

δ ≈ 0.535

δ ≈ 0.129

δ ≈ 0.121

No such δ exists

1 2 3

1. According to the ε - δ definition of the limit, if

and ε = 0.25, there is a value of δ for which |x - 2| < δ implies
|f(x) - 1| < ε. What is this value of δ?

δ = 0.25

δ = 1

δ = 1.25

δ = 2

No such δ exists

2. According to the ε - δ definition of the limit, if

and ε = 2.5, there is a value of δ for which |x - 1| < δ implies |f(x) - 0| < ε. What is this value of δ?

δ = 2.5

δ = 1.5

δ = 1

δ = 0

No such δ exists

3. According to the ε - δ definition of the limit, if

, then for every value of ε > 0, there is a value of δ for which |x - 1| < δ implies |f(x) - 0| < ε. What is the smallest value of ε, below which no such δ exists?

ε = 0

ε = 1

ε = 2

ε = 2.5

ε = 3