The prime number theorem, PNT, states that the prime counting function $\pi(n)$ is asymptotically equivalent to Gauss' first approximation:
$$\pi(n) \sim \frac{n}{\ln(n)}$$
We know this means that
$$\lim_{n \rightarrow \infty}\frac{\pi(n)}{n/\ln(n)} \rightarrow 1$$
Gauss' second approximation is the logarithmic integral $\text{li}(n)$, and this produces better approximations for $\pi(n)$.
$$\pi(n) \sim \text{li}(n) = \int_{0}^{n}\frac{1}{\ln(x)}dx$$
The prime number theorem is also stated in terms of this $\text{li}(n)$.
For the PNT to be valid with both approximations, the two approximations must both be asymptotically equivalent. That is,
$$\text{li}(n) \sim \frac{n}{\ln(n)}$$
Question: How does one prove the two approximations are asymptotically equivalent?
We can expand the logarithmic integral using integration by parts, and this process leaves an integral. Several applications extract several terms of the form $\frac{An}{\ln(n)}$.
$$\text{li}(n) = \frac{n}{\ln(n)} + \frac{n}{\ln^2(n)} + \frac{2n}{\ln^3(n)} + \int_0^n\frac{6}{\ln^4(n)} + C$$
Can we argue that dividing each term by $\frac{n}{\ln(n)}$, and taking the limit $n \rightarrow \infty$, leaves terms that all tend to zero except the first term which tends to 1?
Can we argue that arbitrary applications of integration by parts results in terms that tend to zero, and that the remaining integral is itself smaller because the $\ln(n)$ in the denominator of the integral has higher and higher powers?
Note: I am not mathematically trained so would appreciate responses with minimal assumptions about terminology.
One way to prove it is to use de l'Hôpital's rule. Let $f(x) = \frac{x}{\log{x}}$ and $g(x) = \int_1^x \frac{dt}{\log{t}}$. Then $$ \lim_{x \to \infty}f(x) = \lim_{x\to\infty}g(x)=\infty $$ and therefore $$ \lim_{x \to \infty}\frac{f(x)}{g(x)} $$ is undefined. We can apply de l'Hôpital once and we obtain $$ \lim_{x \to \infty}\frac{f(x)}{g(x)} = \lim_{x \to \infty}\frac{f'(x)}{g'(x)}. $$
We have $$ f'(x) = \frac{\log{x}-1}{\log^2 x} \qquad \text{ and } \qquad g'(x)=\frac{1}{\log{x}}, $$
the latter following from the fundamental theorem of calculus. Finally we have $$ \lim_{x \to \infty}\frac{f(x)}{g(x)} = \lim_{x \to \infty}\frac{f'(x)}{g'(x)} = \frac{(\log x - 1)\log x}{\log^2 x} = 1. $$