## 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

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. viniterranova Group Title

Maybe this is because of the languague usage. The big chanllenge is to translate from the portuguese to English.