I was looking at Stirling's approximation and I came up with the following sort of asymptote:
$$f(x)=g(x+a(x))$$
where $a(x)\to0$ as $x\to\infty$. I noticed this because, if we let $S(x)$ be Stirling's approximation,
$$x!-S(x)\to\infty,x\to\infty$$
But the horizontal difference tends to $0$ ie $x!=S(x+a(x))\implies a(x)\to0$ as $x\to\infty$.
What kind of asymptote is this? Is this equivalent to $\lim_{x\to\infty}\frac{f(x)}{g(x)}=1$?
I'm assuming that $f$ and $g$ tend to infinity as $x$ tends to infinity.
If $g$ is $C^1$ then $$g(x+a(x))=g(x)+g'(x)a(x)+o(a(x))$$ and since $a(x)\rightarrow 0$ as $x$ tends to infinity we can write $$g(x+a(x))=g(x)+g'(a(x))a(x)+o(1)$$ This means that $$f(x)=g(x)+g'(a(x))a(x)+o(1)$$ that is $$\lim_{x\rightarrow\infty}\dfrac{f(x)}{g(x)}=1+\lim_{x\rightarrow\infty}\dfrac{g'(a(x))a(x)}{g(x)}$$ if both sides exist. If $g'$ is continuous at 0 then $g'(a(x))\rightarrow g'(0)$ as $x\rightarrow\infty$. Since $a(x)\rightarrow 0$ and $g(x)\rightarrow\infty$ we get $$\lim_{x\rightarrow\infty}\dfrac{f(x)}{g(x)}=1$$