Preliminary analysis of the problem (guessing a value for 𝛿). Let 𝜀 be a given positive number. We want to find a number 𝛿 such that

if 0 < |x − 6| < 𝛿 then |(2x − 5) − 7| < 𝜀.
But
|(2x − 5) − 7| = |2x − 12| = 2


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.
Therefore, we want 𝛿 such that
if 0 < |x − 6| < 𝛿 then 2


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< 𝜀
that is, if 0 < |x − 6| < 𝛿 then


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<
𝜀
2
.

This suggests that we should choose 𝛿 = 𝜀/2.

1 answer

It looks like you're trying to analyze the problem of finding a suitable \(\delta\) in order to satisfy the condition for a limit, specifically for the function \(f(x) = 2x - 5\) as \(x\) approaches \(6\). The goal is to ensure that \(|f(x) - 7| < \epsilon\) whenever \(|x - 6| < \delta\).

Let's break it down step-by-step:

  1. Determine the expression:

    We want to analyze \(|f(x) - 7|\):

    \[ f(x) = 2x - 5 \] Therefore,

    \[ |f(x) - 7| = |(2x - 5) - 7| = |2x - 12| = 2|x - 6| \]

  2. Set the goal:

    We want this expression to be less than \(\epsilon\):

    \[ |2x - 12| < \epsilon \]

    From our earlier calculation:

    \[ 2|x - 6| < \epsilon \]

  3. Manipulate the inequality:

    To make this easier, we can divide both sides by \(2\):

    \[ |x - 6| < \frac{\epsilon}{2} \]

  4. Choose \(\delta\):

    Thus, we can choose:

    \[ \delta = \frac{\epsilon}{2} \]

  5. Conclusion:

    In summary, if we choose \(\delta = \frac{\epsilon}{2}\), then whenever \(0 < |x - 6| < \delta\), it will guarantee that:

    \[ |(2x - 5) - 7| < \epsilon \]

This procedure is common in proving limits. The approach you've taken is a good way to analyze the continuity of the function at the given point. If you have additional questions or need clarification on this topic, feel free to ask!