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I'm interested to evaluate this $\displaystyle \lim_{x\to \infty } \frac{\tan x}{\log x} $, But really I can't juge whether it is convergent or divergent , Wolfram alpha suggested that dosn't exist as shown here Also l'hopitale rule dosn't work here however $\log x$ is a function of $C^{\infty}$, I think the function $\frac{\tan x}{\log x} $ related to irrationality measure of $\pi$ which it is less clear at all, Really I need a help about evaluation of that limit if it is converge ?

Note I'm interested to the titled limit because the Taylor expansion of $\frac{\tan x}{\log x} $ arround $x=0$ is $\frac{x}{\log x} $ which it does describes Prime counting function , we may need $\frac{\tan x}{\log x} $ to know more about distribution of primes for large integer less than $x$

zeraoulia rafik
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    The limit is divergent. As an informal argument, consider that $\tan x$ periodically jumps to infinity while $\log x$ increases very very slowly, and that $\tan x$ takes on both positive and negative values as $x \to \infty$. – Math2718 Jun 19 '20 at 20:29
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    $\tan $ is not defined at any odd multiple of $\frac12\pi$. – John Bentin Jun 19 '20 at 20:30
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    Of course $\log x$ is a $C^\infty$ function. But you need to check important hypotheses when you try to apply L'Hôpital's rule. – Ted Shifrin Jun 19 '20 at 20:40
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    Since $\tan(x)$ is periodic $\underset{x\to\infty}{\text{lim}}\frac{\tan(x)}{\log(x)}$ is indeterminate. The distribution of the primes is more closely related to $\text{li}(x)$ than $\frac{x}{\log(x)}$. – Steven Clark Jun 19 '20 at 22:09
  • As it has already been said, the limit does not exist. It is woth to plot the function to have a better idea on why is that the case. Have you tried it? You can type, for example, Plot[Tan[x]/Log[x],{x,0,9.5*Pi}] into Wolfram Alpha. – Gary Jun 22 '20 at 09:03

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$\lim_{k \to +\infty} \frac{\tan(\lfloor2k\rfloor\pi)}{\log(\lfloor2k\rfloor\pi))}=0$, since the numerator is always zero. Here $\lfloor a \rfloor$ denotes the integer part of $a$.
Also, $\lim_{k \to +\infty} \frac{\tan(\lfloor2k\rfloor\pi+\pi/2-\frac{1}{k})}{\log(\lfloor2k\rfloor\pi)+\pi/2-\frac{1}{k})}=+\infty$. To see this, observe that $\tan x$ is a periodic function which close (from the left) to $\pi/2$ is greater than $x$. See here why $\tan x>x, 0<x<\pi/2$. We also know that $\frac{x}{\log x}\to +\infty$.
So, for $x$ defined as in the second line, $\lim_{x \to +\infty} \frac{\tan x}{\log x}=+\infty$.
But for $x$ defined as in the first line, $\lim_{x \to +\infty} \frac{\tan x}{\log x}=0$.
This shows Wolfram alpha was right, and that the limit does not exist.

Konstantinos Gaitanas
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