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viniterranova
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Usint the epsilon delta definition evaluate limit x>3 x^2=9, where epsilon = 0,005.
 2 years ago
 2 years ago
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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yakeyglee Group TitleBest ResponseYou've already chosen the best response.0
Your question doesn't really make sense. The epsilondelta 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 \(xx_0 < \epsilon\) then \(f(x)L<\delta\). The epsilondelta definition of the limit is only effective if you consider all \(\epsilon\), not a particular value of \(\epsilon\).
 2 years ago

viniterranova Group TitleBest ResponseYou've already chosen the best response.0
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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