What is the asymptote of a function bounded by two functions with the same limit at $+\infty$? Say we have $$A(x)f(x)<g(x)<B(x)f(x), \hspace{0.5cm} \mbox{$\forall x>x_0.$} $$ Suppose that $$\lim_{x \to +\infty} A(x)=\lim_{x \to +\infty} B(x)=1.$$ Then is it both true that $g(x)\sim f(x)$, i.e., $g(x)$ is asymptotic to $f(x)$, and $\lim\limits_{x \to +\infty} g(x)=f(x)$? I used the websites below to research this, but was unable to come away with anything meaningful.
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$\begingroup$ The notation $\lim_{x \to \infty} g(x)=f(x)$ is meaningless, since the left-hand side doesn't depend on $x$, while the right-hand side does. $\endgroup$– Hans LundmarkJul 8, 2021 at 15:53
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By your assumption $\lim_{x\to\infty}A(x)=\lim_{x\to\infty}B(x)=1$, for every $\epsilon\in(0,1)$, there exists $x_1>x_0$ such that $$1+\epsilon>A(x)>1-\epsilon>0,\quad 1+\epsilon>B(x)>1-\epsilon>0,\quad \forall x>x_1.$$ This means that $f(x)\neq 0$ and $g(x)\neq 0$ for all $x>x_1$, otherwise the relation $A(x)f(x)<g(x)<B(x)f(x)$ fails at those points with $f(x)=0$ or $g(x)=0$. With this being said, the quotient $\frac{g(x)}{f(x)}$ is well defined for $x>x_1$. Moreover, we have $$A(x)<\frac{g(x)}{f(x)}<B(x)\quad\text{or}\quad A(x)>\frac{g(x)}{f(x)}>B(x)$$ for $x>x_1$, depending on the sign of $f(x)$. It follows that $$1-\epsilon<\frac{g(x)}{f(x)}<1+\epsilon,\quad\forall x>x_1,$$ which yields $$\lim_{x\to \infty}\frac{g(x)}{f(x)}=1.$$ So $g$ and $f$ have the same asymptotic behavior when $x\to\infty$.
However, this doesn't necessarily imply that $\lim_{x\to\infty}g(x)=\lim_{x\to\infty}f(x)$. A simple example is $$f(x)=g(x)=|\sin(x)|+1,$$ in which case both functions have no limit when $x\to\infty$.
