Using the \(\epsilon \)-\(\delta \) Definition of a Limit

Prove that if \(\lim\limits_{x\rightarrow c}f(x)\gt0\), then there is an open interval around \(c\), for which \(f(x)\gt0\) everywhere in the interval except possibly at \(c\).

Solution Suppose \(\lim\limits_{x\rightarrow c}f(x)=L\gt0\). Then given any \(\epsilon \gt0\), there is a \(\delta \gt0\) so that \begin{equation*} \hbox{whenever }\quad 0\lt\vert x-c\vert \lt\delta \qquad \hbox{ then}\quad \vert f(x)-L\vert \lt\epsilon \ \end{equation*}

If \(\epsilon =\dfrac{L}{2}\), then from the definition of limit, there is a \(\delta \gt0\) so that \begin{equation*} \hbox{whenever } 0\lt\vert x-c\vert \lt\delta\quad \hbox{ then } \vert f(x)-L\vert \lt\dfrac{L}{2}\quad \hbox{ or equivalently, } \dfrac{L}{2}\lt f(x)\lt\dfrac{3L}{2} \end{equation*} Since \(\dfrac{L}{2}\gt0\), the last statement proves our assertion that \(f(x)\gt0\) for all \(x\) in the interval \((c-\delta ,c+\delta)\).