How can we prove that $\lim_{x\to\infty}\left(\frac{\pi (x)}{x}-\frac{1}{\log x}\right)=0$?

Question

The prime number theorem states that $$\lim_{x\to\infty}\frac{\pi (x)}{x/\log x}=1$$ where $\pi$ is the prime counting function (https://en.wikipedia.org/wiki/Prime-counting_function). In a YouTube video (https://youtu.be/zlm1aajH6gY at 4:19) they essentially labeled the following as "Gauss's conjecture": $$\lim_{x\to\infty}\left(\frac{\pi (x)}{x}-\frac{1}{\log x}\right)=0.$$ I don't understand at all why/if Gauss's conjecture follows from the prime number theorem. The prime number theorem is a statement about a ratio, whereas Gauss's conjecture is a statement about a difference.

It's very confusing as we know that the prime number theorem does not imply $$\lim_{x\to\infty}\left(\pi (x)-\frac{x}{\log x}\right)=0,$$ so why is Gauss's conjecture true?

Answer 1

In general, if we have $f(x) \sim g(x)$ and $g(x) \to 0$ as $x\to \infty$, then we also have $f(x) - g(x) \to 0$.

The proof is simple:

If $f(x) \sim g(x)$, then $\frac{f(x)}{g(x)} \to 1$, and therefore $\frac{f(x)-g(x)}{g(x)} \to 0$. Then $$f(x) - g(x) = g(x)\cdot \frac{f(x)-g(x)}{g(x)} \to 0\cdot 0 = 0$$

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In your case, $g(x) = 1/\log x$ does satisfy $1/\log(x) \to 0$ as $x\to \infty$, so we can conclude from $\frac{\pi(x)}{x} \sim \frac{1}{\log x}$ that their difference tends to $0$. It doesn't work with $\frac{x}{\log x}$ (unfortunately) since $\frac{x}{\log x} \to \infty$.

Finally, note that we can expand the above theorem to the case when $\limsup_{x\to\infty} g(x) < \infty$ instead of $\lim_{x\to\infty} g(x) = 0$, but we don't gain anything in this context from this generalization.

Answer 2

$$\lim_{x\to\infty}\frac{\pi (x)}{x/\log x}=1,$$ means that $\pi(x)\sim \frac{x}{\log x}$ when $x\to\infty$, so $\frac{\pi(x)}{x}\sim \frac{1}{\log x}$, which means that $$\frac{\pi(x)}{x}-\frac{1}{\log x}\to 0,$$ as $x\to \infty$, since $\frac{1}{\log x}\to 0$ as $x\to \infty$

Answer 3

$$\lim\frac{\pi(x)}{x/\log x}=1$$ can be written as:

$$\frac{\pi(x)\log x}{x}=1+o(1)$$ So $$\frac{\pi(x)}{x}=\frac1{\log x}+o\left(\frac1{\log x}\right)$$

Hence:

$$\frac{\pi(x)}x-\frac1{\log x}=o\left(\frac1{\log x}\right)\to 0.$$

Answer 4

Notice that the expression

$$\frac{\pi (x)}{x}-\frac{1}{\log x}\ =\ \frac{1}{\log x}\left(\frac{\pi (x)}{x/\log x} -\ 1\right).$$

Since you already know that

$$\lim_{x\to\infty}{\left(\frac{\pi (x)}{x/\log x} - 1\right)} = 0$$

by the Prime Number Theorem, Gauss's conjecture follows. It's also worth noting that even though RHS of the product tends to zero in the limit, you also have the $\frac{1}{\log x}$ term meaning the much weaker

$$\lim_{x\to\infty}{\frac{\pi(x)}{x}}\ =\ 0$$

would still yield the same result.