Calculus Tutorial 2.9.004.

 
Problem Statement

{3,5,7,9}
eval rand(2,9)
eval rand(1,9)
eval 7*3-4
eval 4 + 17
eval rand(1,9)/1000.
eval 7*3
eval round(0.007/3,6)

Consider , where f(x) = 3 x - 4. Show that |f(x) - 17| < 3 δ if |x -7| < δ.

Find δ such that |f(x) - 17| < 0.007 if |x - 7| < δ.

Prove rigorously that .

 
Step 1

To show that |f(x) - 17| < 3 δ if |x - 7| < δ, we substitute f(x) = 3 x - 4 into the expression |f(x) - 17| and algebraically simplify.

Question Sequence

Question 1

|f(x) - 17| = |(3 x - 4) - 17|

= |3 x - 21|

= |x - 7|

Incorrect.
Correct.

 
Step 2

Question Sequence

Find a δ such that |f(x) - 17| < 0.007 if |x - 7| < δ. In Step 1 we showed that |f(x) - 17| < 3 δ if |x - 7| < δ.

Question 3

Therefore we can set 3 δ = 0.007 and solve for δ. This will give us the maximum δ that satisfies the conditions. (Round your answer to six decimal places.)

3 δ = 0.007

δ =

Incorrect.
Correct.

 
Step 3

Prove rigorously that .

Question 4

Recall the formal definition of a limit. Suppose that f(x) is defined for all x in an open interval containing c (but not necessarily at x = c). Then if for all ε > 0, there exists δ > 0 such that |f(x) - L| < ε if 0 < |x - c| < δ.

According to this definition, to prove that , we must show that for all ε > 0, there exists δ > 0 such that |f(x) - | < ε if 0 < |x - | < δ.

Incorrect.
Correct.

 
Step 4

Algebraically manipulate |f(x) - 17| so that we get a factor of |x - 7|. Then solve for δ in terms of ε.

|f(x) - 17| = 3 |x - 7| < ε

We see the gap between f(x) and 17 is 3 times the gap between x and 7, which we wish to call δ. Since we also wish to keep |f(x) − 17| < ε, we need to choose δ in terms of ε so that |x − 7| < δ and this equation holds. We do this by choosing the following value of δ.

Question 5

δ = ε/

Correct.
Incorrect.

This value of δ satisfies the formal definition of limit and proves .