The steps you've laid out follow the formal epsilon-delta definition of a limit. However, there seems to be a small mistake in the part where you determined \( |(2x - 5) - 7| \).
Let's clarify the proof step by step:
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Goal: We want to show that \[ \lim_{x \to 6} (2x - 5) = 7. \]
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Epsilon-Delta Definition: According to the epsilon-delta definition, for every \( \epsilon > 0 \), we must find a \( \delta > 0 \) such that if \( 0 < |x - 6| < \delta \), then it follows that \( |(2x - 5) - 7| < \epsilon \).
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Set up the Expression:
- First, simplify \( |(2x - 5) - 7| \): \[ |(2x - 5) - 7| = |2x - 12| = 2|x - 6|. \]
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Choose \( \delta \): Set \( \delta = \frac{\epsilon}{2} \).
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Show the Implication: If \( 0 < |x - 6| < \delta \), then: \[ |2x - 12| = 2|x - 6| < 2\delta. \] Substituting \( \delta = \frac{\epsilon}{2} \), \[ 2\delta = 2 \cdot \frac{\epsilon}{2} = \epsilon. \] Thus, \[ |2x - 12| < \epsilon. \]
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Conclude: Therefore, for every \( \epsilon > 0 \), choosing \( \delta = \frac{\epsilon}{2} \) ensures that if \( 0 < |x - 6| < \delta \), then \( |(2x - 5) - 7| < \epsilon \). Hence, we conclude that: \[ \lim_{x \to 6} (2x - 5) = 7. \]
Your original proof was on the right track, but itβs important to ensure the transition steps are explicit to prevent confusion. Just remember to keep track of how \( |2x - 12| \) relates directly to \( |x - 6| \) and how choosing \( \delta \) influences this relationship.
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