$\epsilon$ $\delta$ again!

Question

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

$\blacksquare$

Answer 1

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.

Answer 2

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 $|f(x)-L|<\dfrac{L}{2}$