$\epsilon$ $\delta$ again! Is the following proof correct?
Theorem. If $\lim_{x\to c}f(x)=L$ and $L>0$ then there is some $\delta>0$ such that for all $x$ such that $0<|x-c|<\delta$, $f(x)>0$.
Proof. Let $\epsilon=\frac{L}{2}$ and since $\lim_{x\to c}f(x)=L$ it follows that for some $\delta>0$ it is the case that 
$$\forall x(0<|x-c|<\delta\implies |f(x)-L|<\epsilon)\tag{1}$$
It is now apparent that given any arbitrary $x$ such that $0<|x-c|<\delta$ it follows that $|f(x)-L|<\epsilon$.
consequently $-\frac{L}{2}<f(x)-L<\frac{L}{2}$ and so $\frac{L}{2}<f(x)<\frac{3L}{2}$ but $0<L$ and therefore $0<\frac{L}{2}$ an thus $0<\frac{L}{2}<f(x)$
$\blacksquare$

May I add that if $\varepsilon =L$ there exist $\delta '$ according to the définition.
so $\forall x\in (c-\delta', c+\delta') \implies 0<f(x)<2L\qquad$ This one is the greater delta.
And for any $\delta \in (0,\delta'],\;|x-c|<\delta \implies f(x)>0$
If I call $\delta_1$ choosen according to $\varepsilon_1=\dfrac{L}{2}$ as you have shown above, we can say this $\delta_1\in (0,\delta']$ and any $\delta\leq\delta_1$,  ($\delta> 0$) are suitable to satisfy $|f(x)-L|<\dfrac{L}{2}$
