# Contradiction proof for a limit law $f(x) \le g(x)$

Suppose that $f(x) \le g(x)$ for all $x$. Prove that $\displaystyle \lim_{x \to a} f(x) \le \lim_{x \to a} g(x)$, provided these limits exist.

I posted a similar question, but this is a different approach

Let $\displaystyle \lim_{x \to a} f(x) = L$

Let $\displaystyle \lim_{x \to a} g(x) = M$

Assume $L > M$

Assume without loss of generality,

$x_1 < x_2 < x_3 < ... < x_{n-1} < x_n = a$

Let $\Delta(x_l) = dx_l$ and infinitely small change. So that $f(x_{n-1})$ exists. Now, Suppose $f(x_{n-1})$ exists.

We know then, $f(x_{n-1}) + dx_l = L$

Assume without loss of generality,

$x_1 < x_2 < x_3 < ... < x_{n-1} < x_n = a$

Let $\Delta(x_l) = dx_l$. So that $g(x_{n-1})$ exists. Now, Suppose $g(x_{n-1})$ exists.

We know then, $g(x_{n-1}) + dx_l = M$

We know $f(x) \le g(x)$

We assumed $L > M \implies f(x_{n-1}) + dx_l > g(x_{n-1}) + dx_l$

This gives us:

$f(x_{n-1}) > g(x_{n-1})$

A Contradiction, which completes the proof.

Suppose that $a$ is a real number, and that $I$ is an open interval which contains $a$, and that $f,g$ are real value functions defined everywhere on I except probably at . If $f$ and $g$ have limits as $x$ approaches $a$ and $f(x) \leq g(x)$ for all $x$ in I\{a}, then $\lim_{x\rightarrow a} f(x) \leq \lim_{x\rightarrow a} g(x)$.

Proof:

Let $h(x) = g(x) - f(x)$. Then $h(x) \geq 0$ for all $x$ in I\{a}.

Then $\lim_{x\rightarrow a} h(x) = \lim_{x\rightarrow a} (g-f)(x)$ = $\lim_{x\rightarrow a} g(x) - \lim_{x\rightarrow a} f(x) = C$

Then it is enough to prove that $C \geq 0$.

If $C < 0$, then there exists $\delta > 0$ for all $x$ in $I$ except possibly at $a$,

such that $0 < |x-a| < \delta$ implies $|h(x) - L| < \frac{|C|}{2}$ implies $h(x) < C + \frac{|C|}{2}$ = $C- \frac{C}{2} = \frac{C}{2} < 0$.

But this contradicts, $h(x) \geq 0$ for all $x$ in I\{a}.

Therefore, $C\geq 0$. Thus, $\lim_{x\rightarrow a} f(x) \leq \lim_{x\rightarrow a} g(x)$.

• is my proof fine? – Amad27 Nov 27 '14 at 7:59