If $f(x)=O(g(x))$ & $g(x)=O(f(x))$ then can we write $f(x)\sim g(x)$ or any other one-line relation? $f(x)=O(g(x))$ & $g(x)=O(f(x))$ then can we write $f(x)\sim g(x)$ or $$\lim_{x \to \infty}[f(x)/g(x)]=1$$ where f(x)=pi(x)(prime counting function) and g(x)=li(x)(logarithmic integral:li_x=Li_x-li_2)
 A: The answer is no, generally speaking. For instance, if $f(x) = x$ and $g(x) = 2x$ then one has $f(x) = O(g(x))$ and $g(x)=O(f(x))$ but $$\lim_{x\to\infty} \frac{g(x)}{f(x)} = 2.$$
Actually, it is also possible that there is no limit at all! For an example, take somthing like $f(x) = 2+\cos(x)$ and $g(x)=1$.
However, there is another notation corresponding to this case: $f(x) \asymp g(x)$, meaning that $f$ and $g$ have the same order of magnitude. An alternative definition is that
$$
0 < \liminf \frac{f(x)}{g(x)} \leq \limsup \frac{f(x)}{g(x)} < \infty.
$$
Notice that, like $\sim$, the notation $\asymp$ defines an equivalence relation.
Concerning you edit. In this special case, it is true that $\pi(x) \sim \mathrm{Li}(x)$ but this is a hard result known as the Prime Number Theorem (see there for a discussion on this subject).
A: It won't necessarily be true that $\lim_{x\to\infty} f(x)/g(x) = 1$, but it will be bounded.
For example, if $f(x) = 3x^2$ and $g(x) = x^2 + 1$, then the limit is $3$. And I think you could even come up with examples where the limit doesn't exist.
In general, $f(x) = O(g(x))$ means
$$
f(x) \leq C_1 g(x)
$$
(for $x \geq x_1$ for some $x_1$) and $g(x) = O(f(x))$ means
$$
g(x) \leq C_2 f(x)
$$
(similarly, for $x \geq x_2$)
so combining these, we have
$$
C_3 g(x) \leq f(x) \leq C_1 g(x)
$$
for $x \geq x_0 := \min(x_1,x_2)$ (and here $C_3 = \dfrac{1}{C_2}$) so we have
$$
C_3 \leq \frac{f(x)}{g(x)} \leq C_1
$$
and from this, we can conclude the limit is bounded.
