# What is the asymptote of a function bounded by two functions with the same limit at $+\infty$?

What is the asymptote of a function bounded by two functions with the same limit at $$+\infty$$? Say we have $$A(x)f(x)x_0.}$$ 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.

https://en.wikipedia.org/wiki/Asymptote

https://en.wikipedia.org/wiki/Limit_of_a_function

• 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. Jul 8, 2021 at 15:53

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 $$A(x)f(x) fails at those points with $$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)}\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$$.