• Mar 3rd 2006, 11:08 AM
if
$\lim_{x\to \infty}\frac{g(x)}{f(x)} = 1$
and
$\lim_{x\to \infty}\frac{h(x)}{f(x)} = 1$
then does this imply that
$\lim_{x\to \infty}\frac{g(x)}{h(x)} = 1$ also?
• Mar 3rd 2006, 11:48 AM
ThePerfectHacker
Quote:

if
$\lim_{x\to \infty}\frac{g(x)}{f(x)} = 1$
and
$\lim_{x\to \infty}\frac{h(x)}{f(x)} = 1$
then does this imply that
$\lim_{x\to \infty}\frac{g(x)}{h(x)} = 1$ also?

If $\frac{h(x)}{f(x)}>0$ then,
$\left(\lim_{x\to\infty}\frac{h(x)}{f(x)}\right)^{-1}=\lim_{x\to\infty}\frac{f(x)}{h(x)}$ because the function $\frac{1}{x}$ is countinous for $x>0$. Since,
$\lim_{x\to\infty}\frac{h(x)}{f(x)}=1$
Then,
$\lim_{x\to\infty}\frac{f(x)}{h(x)}=1$
Then,
$\lim_{x\to\infty}\frac{f(x)}{h(x)}\lim_{x\to\infty }\frac{g(x)}{f(x)}=1$
Since the limits exists,
$\lim_{x\to\infty}\frac{\not{f(x)}}{h(x)}\frac{g(x) }{\not{f(x)}}=1$

Note: I only proved this for a specific case when that fraction is always positive. I do not know if it is true in a general case, but it might.
• Mar 3rd 2006, 01:48 PM