Given the condition:
$0\leq \frac{\psi(x)}{x}-\frac{\vartheta(x)}{x}\leq\frac{(\log x)^2}{2\sqrt{x}\log(2)} ,$
how to prove the formula in the middle tend to zero as limit of x goes to infinite
Given the condition:
$0\leq \frac{\psi(x)}{x}-\frac{\vartheta(x)}{x}\leq\frac{(\log x)^2}{2\sqrt{x}\log(2)} ,$
how to prove the formula in the middle tend to zero as limit of x goes to infinite
There is a general, highly intuitive idea at play here. This is called the squeeze theorem (wiki link). The idea is that if we know that our function is always bounded by something above, and by another thing below, then when the two bounds take the same value, so does our function. This should feel right. Draw a picture to illustrate.
No matter how strange the behavior of the blue line, as long as we know its bounded by green and red, we can say some things about its behavior. This picture describes the case when the bounds meet at the same point at $x = 0$. Even though the problem we are considering is for $x \to \infty$, the idea is the same.
It turns out that it is well known that $\dfrac{\log ^2 x}{\sqrt x \log 2} \to 0$ as $x \to \infty$. In fact, $\dfrac{\log^p x}{x^{0 + \epsilon}}$ for any $p$ and any $\epsilon > 0$ also tends to $0$. In words, this means that logs grow really, really, really slowly compared to any polynomial, or even roots. This follows from L'Hopital's rule or Taylor expansions, or many other methods. But the important part is that it goes to $0$.
Then we have two bounds: $0$ and something that goes to $0$. Thus our function must also tend to $0$.