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 $(c-\delta ,c+\delta)$.