$\epsilon$ $\delta$ 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 $0<\frac{L}{2}<f(x)$
$\blacksquare$

It is correct, but too long, since you write$$\forall x(0<|x-c|<\delta\implies |f(x)-L|<\epsilon)$$and, right after that, “It is now apparent that given any arbitrary $x$ such that $0<|x-c|<\delta$ it follows that $|f(x)-L|<\epsilon$.” Therefore, you wrote the same thing twice.
Besides, after making $\epsilon=\frac L2$, there is no need to use the symbol $\epsilon$ again. But using it is not an error, of course.
