Asymptotic equality and inequalities

Question

If $f$ equals $g$ asymptotically, i.e., $f(x)/g(x) \to 1$ (as $x \to \infty$), and $h \leq f$, does that mean $h \leq g$ for sufficiently large $x$? It seems to be true because the relative error, $|f/g - 1|$, tends to zero. I, however, think it is not true in general, that is, we cannot always replace $f$ with $g$ in inequalities. Am I correct?

Answer 1

It is false: Consider $f\colon x\mapsto x+1$, $h=f$ and $g=f-1$. You do have $\frac{f}{g}\to 1$ and $h \leq f$, but always $h > g$.

However, if you have $h$ multiplicatively bounded away from $f$ (that is, there exists $\epsilon > 0$, $A\in \mathbb R$ such that $h(x) \leq f(x)(1 - \epsilon)$ for all $x\geq A$), then what you want holds.

Answer 2

Simple counterexample: take $f(x)=x^2+1, \ g(x)=x^2, \ h(x)=x^2+0.5$