Find d for |f(x) - (1/5)| < .05 if |x-5| < d

[Soaring Crane]

A positive number E and the limit L of a function f at a are given. Find a number d such that |f(x) - L| < E if 0 < |x-a| < d.

Lim x --> 5 (1/x) = 1/5; E = 0.05

The answer or value of d = 1/505.

These are the steps that I did, and, unfortunately, my efforts did not end in that result.

|f(x) - (1/5)| < .05 if |x-5| < d.

a. f(5 - d_1) = (1/5) + .05 = 0.25

1/(5 - d_1) = 0.25

d_1 = 1

b. f(5 + d_2) = (1/5) - 0.05 = .15

1/(5 + d_2) = 0.15

d_2 = 1.67

Neither delta fits the answer. Why is my proof terribly incorrect? Is there an alternate method?

Thanks.

 

[member 553083]

Your proof is not incorrect, but you are using two different delta values. For |f(x) - L| < E to be true, 0 < |x-a| < d must be true for the same value of d. You need to set a single value for d that satisfies both of these conditions. To do this, start by setting d to the smaller of the two deltas (in this case, d = 1). Then, check that |f(5+d) - L| < E is also true. If it is not, then increase the value of d until it is. In this example, we have |f(5+1) - L| = |(1/6) - (1/5)| = 0.16, which is not less than 0.05, so we need to increase the value of d. We can do this by setting d = 1.67 (the larger of the two deltas you calculated), which gives us |f(5+1.67) - L| = |(1/6.67) - (1/5)| = 0.04, which is less than 0.05. Therefore, the answer is d = 1.67.