How does adding o(h) make a statement more precise? From the Introduction to Probability Models 12th Edition by Sheldon M Ross, there exists a paragraph within Section 5.3.2 - Definition of the Poisson Process

The $o(h)$ notation can be used to make statements more precise. For instance, if
$X$ is continuous with density $f$ and failure rate function $\lambda(t)$, then the approximate statements
$$ P(t <X <t +h) \approx f (t)h$$
$$ P(t <X <t +h | X>t) ≈ \lambda (t) h$$
can be precisely expressed as
$$P(t <X <t +h) = f (t)h+o(h)$$
$$P(t <X <t +h|X>t) = \lambda(t)h+o(h)$$

How does this make it more precise, I understand $o(h)$ is just a placeholder for any function such that
$$\lim\limits_{h \to 0} \frac{f(h)}{h} = 0$$
However, how does this make the actual expression more precise? Isn't it still the same equation as before?
 A: There are 2 common ways to make statements Precise.
(1) $P \approx 1000$ is not Precise. Is $P=1012$ ok ? Is $P=1001$ ok ? Maybe it means $P=1000.1$ ? Maybe it is means more strictly $P=1000.0001$ ?
We can make it Precise by stating the bounds :
$P = 1000 (+|-) 0.01 $
This means :
$1000 -0.01 \le P \le 1000 + 0.01 $
(2) $P(x) \approx 2x^3$ is not Precise. What term is missing ? What comparisons can be made with other related values ? How close is the equality ? How can we use it in other Equations ?
We can make it Precise by stating the including the missing term :
(2A) $P(x)=2x^3+o(x^2)$
In context, it means :
$P(x)=2x^3+ax^2+bx+c$
When required, we may try to evaluate $a,b,c$.
(2B) $P(x)=2x^3+o(x)$
This may mean :
$P(x)=2x^3+ax+b$
When required, we may try to compare with $Q(x)=2x^3+\sqrt x$.
(2C) $P(x)=2x^3+o(1)$
This means :
$P(x)=2x^3+a$
When required, we can take Derivative and eliminate the Constant.
(2D) $P(x)=2x^3+o(\sqrt x)$
Meaning :
$P(x)=2x^3+a\sqrt x$
When required, we can use this in $P(x)-Q(x)$ & manipulate it.
All these are $P(x) \approx 2x^3$ but with lot of variety.
We are unable to use $P(x) \approx 2x^3$ in such ways because that is not Precise.
