Determine x1 and x2 in Terms of ε: A Step‑by‑Step Guide
The phrase “determine x₁ and x₂ in terms of ε” most often appears in calculus when we talk about ε‑neighborhoods around a point. In an ε‑δ proof we need to find an interval ((x_1, x_2)) such that whenever a variable stays inside this interval, a certain condition (often a limit) is satisfied. Understanding how to express the endpoints (x_1) and (x_2) directly in terms of ε is a fundamental skill for anyone studying analysis, and it also shows up in engineering, physics, and computer science when we need to bound errors or tolerances Nothing fancy..
This is the bit that actually matters in practice.
Below you will find a complete, easy‑to‑follow explanation that covers the concept, the algebra, and the reasoning behind it. The article is organized with clear headings, bolded key points, and practical examples so you can apply the method to any problem that asks you to “determine x₁ and x₂ in terms of ε.”
Introduction
When a mathematician writes “(0 < |x - a| < \varepsilon)”, they are describing all points (x) that lie within a distance (\varepsilon) of a fixed point (a) on the real line. The set of such points is an open interval ((a-\varepsilon,;a+\varepsilon)). In many textbooks this interval is denoted by ((x_1, x_2)), where
Quick note before moving on.
[ x_1 = a - \varepsilon,\qquad x_2 = a + \varepsilon . ]
Finding (x_1) and (x_2) in terms of ε is simply the process of rewriting the inequality in interval notation. This skill becomes especially useful when you need to prove limits, continuity, or differentiability using the ε‑δ definition. The opening paragraph above also serves as a meta description that includes the main keyword determine x1 and x2 in terms of ε for SEO purposes Simple as that..
Understanding ε‑Neighborhoods
What is an ε‑neighborhood?
An ε‑neighborhood of a point (a) is the collection of all real numbers that are less than ε away from (a). Which means visually, it looks like a line segment centered at (a) with radius ε. Because the inequality is strict ((|x-a| < \varepsilon)), the endpoints themselves are not included in the set. If we wanted to include the endpoints, we would use a non‑strict inequality ((\le)), which would give a closed interval ([a-\varepsilon,;a+\varepsilon]) It's one of those things that adds up..
Why does this matter?
In analysis, we often need to show that we can make a function’s output arbitrarily close to a limit by forcing the input to stay within a tiny interval around a point. The interval’s endpoints are exactly the (x_1) and (x_2) we are asked to determine. By expressing them directly in terms of ε, we make the relationship between the “closeness” of the input and the desired precision crystal clear Small thing, real impact..
Steps to Determine x₁ and x₂ in Terms of ε
Below is a step‑by‑step recipe you can follow for any problem that asks you to find the interval ((x_1, x_2)) defined by (|x-a| < \varepsilon).
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Identify the center point (a).
This is the number that appears inside the absolute value. Take this: in (|x-5| < 0.1), the center is (a = 5). -
Write the inequality without the absolute value.
The statement (|x-a| < \varepsilon) is equivalent to two simultaneous inequalities:[ -\varepsilon < x - a < \varepsilon . ]
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Add (a) to all three parts.
This isolates (x):[ a - \varepsilon < x < a + \varepsilon . ]
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Read off the endpoints.
The left endpoint is (x_1 = a - \varepsilon); the right endpoint is (x_2 = a + \varepsilon) No workaround needed.. -
State the interval.
The solution set is ((x_1, x_2) = (a-\varepsilon,;a+\varepsilon)). If the problem asks for a closed interval, replace the inequality signs with (\le) and keep the same endpoints.
Quick Checklist
- Center identified? ✔️
- Inequality split correctly? ✔️
- Addition performed? ✔️
- Endpoints written in terms of ε? ✔️
If any step fails, double‑check the algebra; the process is straightforward and rarely contains hidden tricks Simple, but easy to overlook..
Example 1: Simple Linear Case
Problem: Determine (x_1) and (x_2) for the inequality (|x - 7| < 0.05) Surprisingly effective..
Solution:
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Center (a = 7) Took long enough..
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Split: (-0.05 < x - 7 < 0.05).
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Add 7: (6.95 < x < 7.05) And it works..
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Endpoints:
[ x_1 = 7 - 0.05 =
Continuing the example, we compute the two endpoints directly from the center (a=7) and the given tolerance (\varepsilon =0.05):
[ x_{1}=7-\varepsilon =7-0.On top of that, 05 = 6. 95, \qquad x_{2}=7+\varepsilon =7+0.Still, 05 = 7. 05.
Thus the (\varepsilon)-neighborhood described by (|x-7|<0.05) is the open interval
[ (x_{1},x_{2}) = (6.95,;7.05). ]
Example 2: A Negative Center and a Larger Tolerance
Problem: Find the interval ((x_{1},x_{2})) that satisfies (|x+4| < 0.12) Small thing, real impact. Worth knowing..
Solution:
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Identify the center.
The expression (|x+4|) can be written as (|x-(-4)|), so the center is (a=-4). -
Split the absolute‑value inequality.
[ -0.12 < x - (-4) < 0.12 ;\Longrightarrow; -0.12 < x+4 < 0.12. ] -
Isolate (x).
Subtract (4) from each part (equivalently, add (-4) to each part):
[ -4.12 < x < -3.88. ] -
Read off the endpoints.
[ x_{1}=a-\varepsilon = -4-0.12 = -4.12,\qquad x_{2}=a+\varepsilon = -4+0.12 = -3.88. ] -
State the interval.
[ (x_{1},x_{2}) = (-4.12,;-3.88). ]
Notice that even when the center is negative, the same formula (x_{1}=a-\varepsilon) and (x_{2}=a+\varepsilon) holds without any algebraic surprises Turns out it matters..
Concluding Remarks
The relationship
[ |x-a|<\varepsilon \quad\Longleftrightarrow\quad a-\varepsilon < x < a+\varepsilon ]
provides a precise bridge between the intuitive notion of “(x) is close to (a)” and the algebraic description of the corresponding open interval. By expressing the endpoints explicitly as (x_{1}=a-\varepsilon) and (x_{2}=a+\varepsilon), we make clear how shrinking (\varepsilon) tightens the interval around the center, a fact that underpins definitions of limits, continuity, and differentiability in real analysis.
When the problem calls for a closed neighborhood, the same endpoints are retained, but the inequality signs become non‑strict ((\le)), yielding ([a-\varepsilon,;a+\varepsilon]). Mastery of this simple transformation equips you with a reliable tool for constructing rigorous (\varepsilon)–(\delta) arguments and for interpreting the geometry of closeness in the real line Simple, but easy to overlook..
