# Show that if $f_1(x) \leq f_2(x)$ for all x in some interval (a,b), then $L_1 \leq L_2$

Suppose the limits $L_1 = \lim_{x \rightarrow a^+}f_1(x)$ and $L_2 = \lim_{x \rightarrow a^+}f_2(x)$ exist. Show that if $f_1(x) \leq f_2(x)$ for all x in some interval (a,b), then $L_1 \leq L_2$.

The proof I've seen just shows that since $f_1(x) \leq f_2(x)$ and since $f_1(x) \rightarrow L_1$ and $f_2(x) \rightarrow L_2$ this shows that $L_1 \leq L_2$. But I don't see how this proves it at all. The proof might be using something like squeeze theorem to prove it, but I want to understand how to prove it directly.

Let $g(x) = f_2(x) - f_1(x)$. Then $g(x) \geqslant 0$ for $x \in (a,b)$ and and $\lim_{x \to a+} g(x) = L = L_2 - L_1$.
Assume $L < 0$. For any $\epsilon > 0$ there exists $\delta > 0$ such that $g(x) < L + \epsilon$ if $a < x < a+ \delta$. Choose $\epsilon = -L/2$. This implies $g(x) < L/2 < 0$ if $a < x < a + \delta$.
Hence $f_2(x) < f_1(x)$ at some point in the interval -- a contradiction.
Therefore, $L = L_2 - L_1 \geqslant 0$ and $L_2 \geqslant L_1$.