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How is the tolerance band related to the absolute value inequality?

The tolerance band from a−εa - \varepsilon to a+εa + \varepsilon is equivalent to the absolute value inequality ∣an−a∣<ε|a_n - a| < \varepsilon. Distance less than ε\varepsilon from the limit aa means that the term ana_n lies strictly inside the open band around aa.

Conditions

  • The limit is aa.
  • The tolerance is ε>0\varepsilon > 0.
  • The sequence term is ana_n.

Reasoning, step by step

  1. Define the tolerance band as the interval (a−ε,a+ε)(a - \varepsilon, a + \varepsilon).
  2. State the absolute value inequality ∣an−a∣<ε|a_n - a| < \varepsilon.
  3. Show that ∣an−a∣<ε|a_n - a| < \varepsilon is equivalent to a−ε<an<a+εa - \varepsilon < a_n < a + \varepsilon.
  4. Conclude that membership in the open band is equivalent to the distance being less than ε\varepsilon.

Example

The card 'The tolerance band' states: 'Distance less than ε is equivalent to membership in the open band around a.' with the formula ∣an−a∣<ε  ⟺  a−ε<an<a+ε|a_n-a|<\varepsilon\iff a-\varepsilon<a_n<a+\varepsilon.

Common misconceptions

  • Confusing the open band with a closed interval.
  • Believing that the term can lie on the boundary of the band.

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