Is $\ln(\pi(e^x)) \sim x?$

Question

Q: Is $\ln(\pi(e^x))\sim x?$

$\DeclareMathOperator{\Li}{Li}$ $\pi(x)$ is the prime counting function. Here's my attempt.

I think to prove or disprove this, I have to show whether or not $$ \lim_{x\to \infty} \frac{\ln(\pi(e^x))}{x}=1.$$

Since $$ \ln(\pi(e^x)) \sim \ln(\Li(e^x)) $$ I will substitute that into the limit:

$$ \lim_{x\to \infty} \frac{\ln(\Li(e^x))}{x}. $$

Now I'll use L'Hopitals rule repeatedly because we have some forms of $\frac{\infty}{\infty}.$

$$ \lim_{x\to\infty} \frac{e^x}{x\Li(e^x)}=\lim_{x\to \infty} \frac{e^x}{\Li(e^x)+e^x}=\lim_{x\to \infty} \frac{e^x}{e^x\left(1+\frac{1}{x}\right)}=\lim_{x\to \infty}\frac{1}{1+\frac{1}{x}}=1. $$

Is this correct?

Answer 1

$\pi(x)\sim x/\ln x$ in the sense that $\pi(x)\ln x/x\to1$ as $x\to\infty$. It follows that $\pi(e^x)x/e^x\to1$ as well, and consequently $\ln(\pi(e^x))+\ln x-x\to0$, which means

$${\ln(\pi(e^x))\over x}-{\ln x\over x}\to1$$

Since ${\ln x\over x}\to0$ as $x\to\infty$, the result follows.