I have two functions $f(x)$ and $g(x)$ with the same asymptote, how do I show that one approaches the limit faster than the other? My intuition tells me that $f(x)$ approaches the limit faster than $g(x)$ if
$$ \lim_{x \to \infty}\frac{f'(x)}{g'(x)}=0$$
The functions are
$$\DeclareMathOperator{\cosech}{csch} f(x) = \frac{1}{\sqrt{(a \mathop{\cosech}{x})^2 + 1}}$$
$$ g(x) = \frac{1}{\sqrt{(b/x)^2 + 1}}$$
where $a$ and $b$ are real constants. Both functions have the limit of 1 as $x$ approaches infinity.
Best Answer
If $f(x)$ and $g(x)$ both tend to $a$ as $x\rightarrow\infty$, then L'Hôpital's rule immediately gives
$$\lim_{x\rightarrow\infty}\frac{f(x)-a}{g(x)-a}=\lim_{x\rightarrow\infty}\frac{f'(x)}{g'(x)}=0$$ which is what you want.