$\lim_{x\to c}f(x)-g(x)$ and Asymptotic Equivalence

Question

I'm confused about asymptotic equivalence.

I read on a textbook that, supposing $\lim_{x\to +\infty} f(x)=+\infty$,

$\lim_{x\to +\infty}f(x)-g(x)= l \in \mathbb{R} \implies f(x) \sim g(x) $

Is this true?

If so are there other cases in which the limit $\lim_{x\to c}f(x)-g(x)$ is related to asymptotic equivalence?

Thanks in advice

Answer 1

(I assume the definition of equivalence you start with is $\frac{f}{g}\to 1$).

Since both $f(x)\to\infty$, $g(x)\to\infty$ when $x\to\infty$ (for $g$, this is easy to see given that $f$ goes to infinity and the difference converges to $\ell$), we know that for $x$ big enough both $f(x),g(x) > 0$ (and we can divide by $g(x)$).

Then,

$$\frac{f(x)}{g(x)} = \frac{f(x)-g(x)}{g(x)} + 1 \xrightarrow[x\to\infty]{} 1$$ as the first term has its numerator going to $\ell\in\mathbb{R}$ and the denominator going to $\infty$. This means that $f\operatorname*{\sim}_{c}g$ indeed.

In general, you can actually define the equivalence $f\operatorname*{\sim}_{c}g$ as $f-g = o(g)$ (at $c$). (and this is often a definition used, since it does not rely on a ratio that can involve dividing by $0$)