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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

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Note that the right hand side $\frac{(\log x)^2}{x \sqrt{x} \log 2}$ approaches $0$ as $x \to 0$. (Of course, the left hand side $\to 0$ clearly.) Now apply the sandwich (also called the squeeze) theorem. – Srivatsan Dec 24 '11 at 22:16
@Srivatsan - i am a beginner that without knowing much of limit, please explain about your hint in your answer, thanks in advance – Victor Dec 24 '11 at 22:21
@Srivatsan - Actually it is x goes to infinity instead of x goes to 0 in my book – Victor Dec 24 '11 at 22:36
I am sorry, what I wrote was a typo. I meant to write $x \to \infty$. – Srivatsan Dec 24 '11 at 22:38
@Will: I take a different view than you. Someone is interested in math, and so why wouldn't we encourage him? I would not be nearly so harsh with someone. Instead, I think that if I explain ideas at his level, and point out where additional material can be found so that he can learn, then no one is harmed. Is this not the best place to ask these questions? I also think that people can learn the most when they are a bit out of their comfort zone - I certainly did. But that's just my view. – mixedmath Dec 24 '11 at 23:10
up vote 5 down vote accepted

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.

Squeeze theorem, stolen from wikipedia as well

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 $0$.

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+1 for an explanation tailored for the OP's level. It might be useful for the OP to point out that the picture depicts a case when the variable $x$ approaches a real number, not $\infty$. – Srivatsan Dec 24 '11 at 23:16
That's a good idea. I'll do that. – mixedmath Dec 24 '11 at 23:21
+1 for reminding me to use l'hopital's rule – Victor Dec 24 '11 at 23:31

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