# Can one prove $f \leq g \implies \lim\limits_{x \to a} f \leq \lim\limits_{x \to a} g$ directly?

The proof I found uses $\epsilon = (l - m)/2$ for $l = \lim\limits_{x \to a} f$ and $m = \lim\limits_{x \to a} g$. They also seem to use the same $\delta$ and they arrived at a contradiction.

Is there a way to prove this directly?

Outline of the proof by contradiction to remind the readers on how this statement is usually proved. (mine is a little different from the text, but same idea)

Choose $\epsilon = (l - m)/2$, then for $\delta = \min\{\delta_f,\delta_g\}$ we have $|x - a| < \delta \implies l - \epsilon< f < l + \epsilon$ and $-m - \epsilon< -g < -m + \epsilon$

Adding the inequalities yields $$0 < f - g$$ or $$g < f$$

Since $f \le g$, $h = g - f \ge 0$. Assuming, $\lim_{x \to a} g(x)$ and $\lim_{x \to a} f(x)$ both exist, $L = \lim_{x \to a} h(x)$ exists, and satisfies $L \ge 0$. To see this, note that for $\epsilon > 0$, there exists $\delta > 0$ such that

$$0 < |x - a| < \delta \implies -\epsilon + L < h(x) < L + \epsilon$$

If $L < 0$, taking $\epsilon = |L|$ yields a contradiction.

In order to construct a correct proof, consider the function $h=f-g$. If $\lim f,\lim g$ are well defined at every point (maybe different from $f(a),g(a)$), then $\lim h$ is also well defined. Now we know $h\ge 0$ for all $x$. Then to show $\lim_{x\rightarrow a} h \ge 0$ you need to prove

$$\forall \epsilon, \exists \delta, \forall x:|x-a|<\delta,h(x)>-\epsilon$$ But now this is trivially satisfied.

• I made a typo. It should have been $\epsilon = \frac{l - m}{2}$ – Hawk Jul 1 '13 at 1:57
• No, it would not make any difference. Your goal is to show $l\le m$, not $f<g$(which you know already). – Bombyx mori Jul 1 '13 at 1:58
• No I assumed $l > m$, but then I arrived $g < f$. So that must mean $l \leq m$ – Hawk Jul 1 '13 at 2:18
• Then this is still not a proof you wanted, because it is a proof by contradiction again. – Bombyx mori Jul 1 '13 at 3:10
• Mine or yours? Why do you have $|h(x) - 0|$? Why $0$? – Hawk Jul 1 '13 at 3:14