A Problem Using the Definition of a Limit

[student34]

Homework Statement

If ε = 10, give a value of δ that satisfies |δ - a| where a = 2 and 0 < δ ≤ 1 and also guarantees that|f(x) − 1/4|< ε where f(x) = 1/(x^2)

Homework Equations

N/A

The Attempt at a Solution

My problem is the solution. The solution is δ = min(1,8) = 1. I am assuming that ε begins at L = 1/4 because |f(x) - L| is usually implied. So I graphed it out; I can see that any position of 0 < δ ≤ 1 from position a works when ε = 10 and 0 < δ ≤ 1. But I have no idea how the answer gets min (1,8). Anything larger than x = 2 is outside of ε.

 

[vela]

Homework Statement

If ε = 10, give a value of δ that satisfies |δ - a| where a = 2 and 0 < δ ≤ 1 and also guarantees that|f(x) − 1/4|< ε where f(x) = 1/(x^2)

What is "satisfies |δ - a|" supposed to mean?

Homework Equations

N/A

The Attempt at a Solution

My problem is the solution. The solution is δ = min(1,8) = 1. I am assuming that ε begins at L = 1/4 because |f(x) - L| is usually implied.

There's no assuming necessary (except perhaps what you mean by "ε begins at L = 1/4"). |f(x) − 1/4|< ε means that f(x) is within ε of 1/4.

So I graphed it out; I can see that any position of 0 < δ ≤ 1 from position a works when ε = 10 and 0 < δ ≤ 1. But I have no idea how the answer gets min (1,8).

Where did the 8 come from in the solutions? I assume the derivation is shown.

Anything larger than x = 2 is outside of ε.

What is this supposed to mean? It doesn't make sense as you are comparing x values to y values.

 

[student34]

What is "satisfies |δ - a|" supposed to mean?

Damn, I meant "satisfies 0 < δ ≤ 1 from the point a = 2". I came to that from the idea that if |x - 2| > δ and 0 < δ ≤ 1, then 1 < x < 3.

There's no assuming necessary (except perhaps what you mean by "ε begins at L = 1/4"). |f(x) − 1/4|< ε means that f(x) is within ε of 1/4.

Where did the 8 come from in the solutions? I assume the derivation is shown.

What is this supposed to mean? It doesn't make sense as you are comparing x values to y values.

I meant that if f(x) = f(a) = f(2) = 1/4, then any x > 2 puts 1/(x^2) outside of the range of ε = 10 which seems to mean that ε = (1/4, 10+1/4) on the y axis. So my answer is, 0 < δ ≤ 1 because ε ≥ {f(1), f(2)}.

 

[vela]

Remember you have an absolute value, so you want -10 < f(x)-1/4 < 10. For some values of x greater than 2, that condition is satisfied.

Did you see where the 8 in the solutions came from?

 

[student34]

Remember you have an absolute value, so you want -10 < f(x)-1/4 < 10. For some values of x greater than 2, that condition is satisfied.

Did you see where the 8 in the solutions came from?

Ah, after a long time, I finally get it - thanks!