Limit of a bounded function is bounded

Question

I am working on proving the following proposition:

If $a \leq f(x) \leq b$ and $\lim_{x \to c}f(x) = L$ then $a \leq L \leq b$.

Starting out with the epsilon-delta definition of a limit, we quickly find that whenever $|x-c| < \delta$, $$L - \epsilon < f(x) < L + \epsilon$$ I am getting stuck here. How do I combine this inequality with the presupposition that $a \leq f(x) \leq b$? It doesn't seem like it's possible to isolate $L$, $a$, and $b$.

Answer 1

Assume this is false, so $L>b$ or $L<a$, in this case we have:

Forall $\epsilon>0$ exists $\delta>0$ such that $|x-c|<\delta$ implies that $L-\epsilon<f(x)<L+\epsilon$

If $L>b$ choose $\epsilon=L+b+1$ to get $b+1<f(x)<2L+b+1$, this is contradiction.

If $L<a$ choose $\epsilon=-L+a-1$ to get $-2L+a-1<f(x)<a-1$, another contradiction.

Hence $a\le L\le b$