0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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, $\lim_{x\rightarrow a} f(x) \leq \lim_{x\rightarrow a} g(x)$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.