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$\Huge \text{stO}$ $\Huge \text{e}\color{red}{\text{disnimorF}}$ $\Huge \text{Orz}$

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For all $\varepsilon > 0$ there exists a $\delta > 0$ such that if $x \in \mathbb{R}$ and $0<|x-a|<\delta$, then $|f(x)-L|<\varepsilon$.

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