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

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yakeygleeBest 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 \(\epsilon\).
 one year ago

viniterranovaBest 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.
 one year ago
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