using the graph of f(x) = 3x - 1 to work with limits

[rocketgo]

Let f(x) = 3x - 1. Then the limit, as x approaches 1, of f(x) is 2. Use the graph of f(x) to answer the following:

a) Find a number s > 0 such that |f(x) - 2| < 0.1 whenever |x - 1| < s

b) Find a number s(2) > 0 such that |f(x) - 2| < 0.01 whenever |x - 1| < s(2)

c) Let E > 0 be a given. In terms of E, find a number s > 0 such that |f(x) - 2| < E whenever 0 < |x - 1| < s

Any help in how do this problem would be helpfull. I am not looking just for an answer, I want to learn how to solve the problem.

Thanks,
Tim

 

[stapel]

a) Work with what they've given you, and see where it takes you:

. . . . .|(3x - 1) - 2| < 0.1

. . . . .|3x - 1 - 2| < 0.1

. . . . .|3x - 3| < 0.1

. . . . .|3(x - 1)| < 0.1

. . . . .3|x - 1| < 0.1

. . . . .|x - 1| < (1/10)(1/3) = 1/30

So, if s < 1/30, then |x - 1| < 1/30, so 3|x - 1| < 1/10 = 0.1, so |f(x) - 2| < 0.1, as required.

Now follow the same process for part (b), and then generalize (work symbolically) for part (c).

Eliz.