How to prove this theorem with $\epsilon$-$\delta$ definition for limits?

Question

If $f(x)\geq 0$ for $a \lt x \lt x_0$ and $\displaystyle\lim_{x \to {x_0}^-} f(x)$ exists, then $\displaystyle\lim_{x \to {x_0}^-} f(x)\geq 0$.

I have issue for $f(x)$s are in $(L, L + \epsilon)$ when assume $\displaystyle\lim_{x \to {x_0}^-} f(x)=L$. Actually, I cannot guarantee that at least one $f(x)$ is in the interval $(L - \epsilon, L)$.

Answer 1

With these kind of questions I usually try a proof by contradiction.

Let $L=\lim_{x\to x_0^-}f(x)$ and assume that $L<0$. Then, we can write $L=-\alpha$ with $\alpha>0$. Let $\epsilon=\alpha$. By the $\epsilon-\delta$ definition for limits, we know that there exists some $\delta>0$ such that $x\in(x_0-\delta,x_0)$ implies $|f(x)-L|<\alpha$. Per definition of the absolute value, this implies that $$ f(x)-\alpha<L<f(x)+\alpha. $$ So, we see that this gives $$ f(x)<L+\alpha=-\alpha+\alpha=0, $$ which is in contradiction with the fact that $f(x)\geq0$ for all $x\in(a,x_0)$.

We conclude that our assumption that $L<0$ must have been wrong, and therefore that $\lim_{x\to x_0^-}f(x)\geq0$.

Answer 2

Assume that $\lim_{x\to x_0^{-}}f(x)=c<0$. Now given $\varepsilon=-c/2$, there exists some $\delta>0$ such that $x_0-x<\delta$ implies $|f(x)-c|\leq -c/2$. Now we take some $x$ with $x_0-x<\delta$, and note that in particular: $$f(x)-c\leq -c/2\Rightarrow f(x)\leq c/2<0$$ Which is a contradiction.