A Problem Using the Definition of a Limit

[The Student]

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).

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.

 

[daon2]

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).

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.

\(\displaystyle \left| \dfrac{1}{x^2}-\dfrac{1}{4}\right|<10 \iff |(2-x)|\cdot |(2+x)|\dfrac{1}{4x^2} < 10\)

We need to bound \(\displaystyle |2+x|\), so assume \(\displaystyle |x-2|<1\). Then

\(\displaystyle 1<x<3 \iff 3< 2+x<5 \iff 3<|2+x|<5\) (since 2+x is positive) and

\(\displaystyle 1<x<3 \iff 4< 4x^2<36 \iff \dfrac{1}{36} < \dfrac{1}{4x^2} < 1/4\).

So with our assumption

\(\displaystyle |(2-x)|\cdot |(2+x)|\dfrac{1}{4x^2} < |2-x|\cdot \dfrac{5}{4} \overbrace{<}^{Set} 10 \rightarrow |x-2|<8\)

 

[The Student]

\(\displaystyle \left| \dfrac{1}{x^2}-\dfrac{1}{4}\right|<10 \iff |(2-x)|\cdot |(2+x)|\dfrac{1}{4x^2} < 10\)

We need to bound \(\displaystyle |2+x|\), so assume \(\displaystyle |x-2|<1\). Then

\(\displaystyle 1<x<3 \iff 3< 2+x<5 \iff 3<|2+x|<5\) (since 2+x is positive) and

\(\displaystyle 1<x<3 \iff 4< 4x^2<36 \iff \dfrac{1}{36} < \dfrac{1}{4x^2} < 1/4\).

So with our assumption

\(\displaystyle |(2-x)|\cdot |(2+x)|\dfrac{1}{4x^2} < |2-x|\cdot \dfrac{5}{4} \overbrace{<}^{Set} 10 \rightarrow |x-2|<8\)

Ok, I now understand how the answer has a max of 8 - thanks, it feels good to know this. But why isn't the interval for δ (0,1] instead of just 1? I drew the graph for this few times, and I see that {f(1), f(2)} is in ε = (1/4, 10+1/4). So this seems to cover the interval of 0 < δ ≤ 1 from point a = 2. As a test, |δ| = 0.6 implies x = 1.6, and f(1.6) = 0.390624 which is in the interval of ε = (1/4, 10+1/4).

 

[daon2]

Ok, I now understand how the answer has a max of 8 - thanks, it feels good to know this. But why isn't the interval for δ (0,1] instead of just 1? I drew the graph for this few times, and I see that {f(1), f(2)} is in ε = (1/4, 10+1/4). So this seems to cover the interval of 0 < δ ≤ 1 from point a = 2. As a test, |δ| = 0.6 implies x = 1.6, and f(1.6) = 0.390624 which is in the interval of ε = (1/4, 10+1/4).

I'm a little confused. What do you mean by the "interval 1"? 1 is an integer. Epsilon is not an interval, yet you are writing it as an interval. You are looking for a delta such that if x is closer to 2 than delta then f(x) is in the interval (1/4-10, 10+1/4). Any delta that works will answer the question, the solution is not unique.

 

[The Student]

I'm a little confused. What do you mean by the "interval 1"? 1 is an integer. Epsilon is not an interval, yet you are writing it as an interval. You are looking for a delta such that if x is closer to 2 than delta then f(x) is in the interval (1/4-10, 10+1/4). Any delta that works will answer the question, the solution is not unique.

Ooops, I completely had a mental malfunction with the symbols. I misinterpreted δ = min(1,8) to mean the value of δ from point a starting 1 unit away from x = a. Thanks again, I understand the whole thing now.