Question

Consider the following limit statement: limx→7(2x−6)=8

Find the corresponding value of δ when ε=0.001.
δ=−0.0005
δ=7.0005
δ =0.05
δ =0.0005

Answer 1

The correct value of δ when ε = 0.001 for the limit statement limx→7(2x−6)=8 is δ = 0.0005.

How to solve

Epsilon-delta definition of a limit: A limit statement limx→aL=L means that for any positive number ε, there exists a positive number δ such that whenever 0 < |x - a| < δ, we have |f(x) - L| < ε.

Apply the definition to the given problem:

We want to find δ such that whenever 0 < |x - 7| < δ, we have |2x - 6 - 8| = |2x - 14| < ε = 0.001.

Solve the inequality: |2x - 14| < 0.001 Since the absolute value of an expression is its distance from zero, we can write this inequality as: -2(0.001) < 2x - 14 < 2(0.001)

Simplifying, we get: -13.998 < 2x < 14.002

Dividing everything by 2, we obtain: -6.999 < x < 7.001

Choose δ: The above inequality tells us that if we choose any value of δ less than 0.001 (e.g., 0.0005), then any x such that 0 < |x - 7| < δ will satisfy the condition |2x - 14| < ε = 0.001.

Therefore, the correct value of δ when ε = 0.001 is δ = 0.0005.

Answer 2

The corresponding value of δ when ε = 0.001 for the given limit
`\(\lim_{{x \to 7}} (2x - 6) = 8\)` is δ = 0.0005.

The limit statement is:
`\(\lim_{{x \to 7}} (2x - 6) = 8\)`

This implies that for any ε > 0, there exists a δ > 0 such that if 0 < |x - 7| < δ, then |(2x - 6) - 8| < ε.

Given ε = 0.001, we need to find the corresponding value of δ.

Let's solve for δ:

|(2x - 6) - 8| < 0.001

Simplify the absolute value inequality:

|2x - 14| < 0.001

Now, solve for |x - 7|:

`\[|x - 7| = \left|(2x - 14)/(2)\right| < (0.001)/(2)\]`

|x - 7| = |2x - 14| < 0.001

Comparing the inequality |2x - 14| < 0.001 with |x - 7| < δ, we see that δ needs to be less than 0.001 for the statement to hold.

Therefore, the correct corresponding value of δ when ε = 0.001 is δ = 0.0005.

Question:

Consider the following limit statement:
`\lim_(x\rightarrow 7)(2x-6)=8`

Find the corresponding value of δ when ε=0.001.

δ=−0.0005

δ=7.0005

δ =0.05

δ =0.0005