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viniterranova Group Title

Usint the epsilon delta definition evaluate limit x->3 x^2=9, where epsilon = 0,005.

  • 2 years ago
  • 2 years ago

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  1. yakeyglee Group Title
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    Your question doesn't really make sense. The epsilon-delta definition of the limit for a limit as \(x \rightarrow x_0\) says that there exists a real number \(L\), such that for all \(\epsilon>0\), there exists a \(\delta > 0\) such that if \(|x-x_0| < \epsilon\) then \(|f(x)-L|<\delta\). The epsilon-delta definition of the limit is only effective if you consider all \(\epsilon\), not a particular value of \(\epsilon\).

    • 2 years ago
  2. viniterranova Group Title
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    Maybe this is because of the languague usage. The big chanllenge is to translate from the portuguese to English.

    • 2 years ago
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