Theorem: Prime-Counting Function is Theta of x over Logarithm of x

We have: 
:$\map \pi x = \map \Theta {\dfrac x {\ln x} }$
where:
:$\Theta$ is $\Theta$ notation
:$\pi$ is the prime counting function.

Proof:
From Second Chebyshev Function is $\map \Theta x$, there exists real numbers $A, B, x_0 > 0$ such that: 
:$A x \le \map \psi x \le B x$ 
for $x \ge x_0$, where $\psi$ is the second Chebyshev function.
From Bounds for Prime-Counting Function in terms of Second Chebyshev Function, there exists a real function $R : \hointr 2 \infty \to \R$ such that: 
:$\dfrac {\map \psi x} {\ln x} + \map R x \le \map \pi x \le \dfrac {2 \map \psi x} {\ln x} + \sqrt x$
for all real numbers $x \ge 2$, with: 
:$R = \map \OO {\sqrt x \ln x}$
We aim to combine these inequalities to first obtain: 
:$\map \pi x \le \dfrac {C_2 x} {\ln x}$
for sufficiently large $x$ and a real number $C_2 > 0$, so we would obtain: 
:$\map \pi x = \map \OO {\dfrac x {\ln x} }$
We will also aim to obtain: 
:$\dfrac {C_1 x} {\ln x} \le \map \pi x$
for sufficiently large $x$ and a real number $C_1 > 0$, so we obtain: 
:$\dfrac x {\ln x} \le \frac 1 {C_1} \map \pi x$
giving: 
:$\dfrac x {\ln x} = \map \OO {\map \pi x}$
At which point we have: 
:$\map \pi x = \map \Theta {\dfrac x {\ln x} }$
which is the demand.
We have, for $x \ge \max \set {x_0, 2}$: 
:$\dfrac {2 \map \psi x} {\ln x} + \sqrt x \le \dfrac {2 B x} {\ln x} + \sqrt x$
From Order of Natural Logarithm Function, for $x \ge 1$ we have: 
:$\ln x \le 2 \sqrt x$ 
So that: 
:$\sqrt x \le \dfrac {2 x} {\ln x}$
So we obtain, for $x \ge \max \set {x_0, 2}$: 
:$\dfrac {2 \map \psi x} {\ln x} + \sqrt x \le \dfrac {\paren {2 B + 2} x} {\ln x}$
Let:
:$M_1 = \max \set {x_0, 2}$
Then, we have:
:$\map \pi x \le \dfrac {\paren {2 B + 2} x} {\ln x}$
for $x \ge M_1$. 
Setting $C_2 = 2 B + 2 > 0$, we have: 
:$\map \pi x \le \dfrac {C_2 x} {\ln x}$
for $x \ge M_1$. 
Now we aim to find a real number $C_1 > 0$ such that: 
:$\dfrac {C_1 x} {\ln x} \le \map \pi x$
Since, for $x \ge 2$:
:$\dfrac {\map \psi x} {\ln x} + \map R x \le \map \pi x$
with $R = \map \OO {\sqrt x \ln x}$, there exists real numbers $C, x_1 > 0$ such that: 
:$\size {\map R x} \le C \sqrt x \ln x$
for $x \ge x_1 \ge 2$.
That is: 
:$-C \sqrt x \ln x \le \map R x$
for $x \ge x_1$.
So, we have: 
:$\dfrac {\map \psi x} {\ln x} + \map R x \ge \dfrac {A x} {\ln x} - C \sqrt x \ln x$ 
for $x \ge \max \set {x_0, x_1}$.
We can show that we have: 
:$\dfrac {A x} {\ln x} - C \sqrt x \ln x \ge \dfrac {A x} {2 \ln x}$
for sufficiently large $x$. 
Note that this inequality is equivalent to: 
:$\dfrac {A x} {2 \ln x} \ge C \sqrt x \ln x$
That is: 
:$\dfrac {\sqrt x} {\paren {\ln x}^2} \ge \dfrac {2 C} A$
From Order of Natural Logarithm Function, we have: 
:$\ln x \le 8 x^{1/8}$
for $x \ge 1$. 
That is: 
:$\dfrac {\sqrt x} {\paren {\ln x}^2} \ge \dfrac {\sqrt x} {\paren {8 x^{1/8} }^2} = \dfrac 1 {64} x^{1/4}$
So it suffices to take: 
:$\dfrac 1 {64} x^{1/4} \ge \dfrac {2 C} A$
Setting: 
:$M_2 = \max \set {x_0, x_1, \paren {\dfrac {128 C} A}^4}$
we have: 
:$\dfrac {\map \psi x} {\ln x} + \map R x \ge \dfrac {A x} {2 \ln x}$
for $x \ge M_2$. 
That is: 
:$\map \pi x \ge \dfrac {A x} {2 \ln x}$
So, taking:
:$C_1 = \dfrac A 2 > 0$
We have: 
:$\map \pi x \ge d\frac {C_1 x} {\ln x}$
for $x \ge M_2$. 
Let: 
:$M = \max \set {M_1, M_2}$
Then, for $x \ge M$, we have: 
:$\dfrac {C_1 x} {\ln x} \le \map \pi x \le \dfrac {C_2 x} {\ln x}$ 
with $C_1, C_2 > 0$, so: 
:$\map \pi x = \map \Theta {\dfrac x {\ln x} }$
as required.
{{qed}}