TITLE: Contradiction proof for a limit law $f(x) \le g(x)$
QUESTION [0 upvotes]: 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.

REPLY [1 votes]: 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, $\lim_{x\rightarrow a} f(x) \leq \lim_{x\rightarrow a} g(x)$.
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