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This question already has an answer here:

What is $\mathop {\lim }\limits_{n \to \infty } \frac{{\tan (n)}}{n}$?

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marked as duplicate by leonbloy, Simon S, JimmyK4542, Alex M., user228113 Jun 13 '16 at 22:03

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  • $\begingroup$ Have you heard of L'Hôpital's rule? $\endgroup$ – ÍgjøgnumMeg Jun 13 '16 at 20:01
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    $\begingroup$ L'Hopital doesn't help. This limit doesn't exist. $\endgroup$ – Simon S Jun 13 '16 at 20:03
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    $\begingroup$ I hate this kind of drafting. We are meant to realize that $n$ is a positive integer, so it is not as stupid as it looks. $\endgroup$ – almagest Jun 13 '16 at 20:14
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    $\begingroup$ Probably the limit does not exist. It will depend on how well $\pi$ is approximable by rationals. $\endgroup$ – GEdgar Jun 13 '16 at 20:22
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    $\begingroup$ See here math.stackexchange.com/a/159489/312 $\endgroup$ – leonbloy Jun 13 '16 at 20:40
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We will use two theorems, the first theorem is a result by Minkowski:

Given any $\theta \notin \mathbb{Q}$ and $\alpha \notin \mathbb{Z}$ such that $x - \theta y - \alpha = 0$ has no integer solutions.
For any $\epsilon > 0$, there exists infinitely many pairs of integers $p,q$ such that $$|q(p - \theta q - \alpha)| < \frac14\quad\text{ and }\quad |p - \theta q - \alpha| < \epsilon$$

Apply this to $\theta = \frac{1}{\pi}$ and $\alpha = -\frac12$, there are infinitely many pairs of non-zero $p, q$ such that

$$\left|q - \pi\left(p+\frac12\right)\right| < \frac{\pi}{4|q|} $$ Using the fact $$\left|\tan x\right| \le \frac{4}{\pi}|x|\quad\text{ for }\quad |x| < \frac{\pi}{4}$$ We find $$\frac{1}{|\tan q|} = \left|\tan\left(q - \pi\left(p+\frac12\right)\right)\right| \le \frac{4}{\pi}\left(\frac{\pi}{4 |q|}\right) = \frac{1}{|q|} \quad\implies\quad \left|\frac{\tan q}{q}\right| \ge 1$$

This implies there are infinitely many positive integer $n$ such that $\displaystyle\;\frac{|\tan n|}{n} > 1$.

The second theorem is Hurwitz's theorem.

For any $\theta \notin \mathbb{Q}$, there are infinitely many pairs of relatively prime integers $p, q$ such that $$\left|\theta - \frac{p}{q}\right| \le \frac{1}{\sqrt{5}q^2}$$

Apply this to $\theta = \pi$, there are infinitely many pairs of relative prime $p, q$ such that

$$|\tan p| = |\tan(p - \pi q)| \le \frac{4}{\pi\sqrt{5}|q|} \le \frac{4}{\sqrt{5}|p|-1}$$ This implies there is a subsequence of $\displaystyle\;\frac{\tan n}{n}$ converges to $0$ as $n \to \infty$.

Combine these two results, we can deduce $\displaystyle\;\lim_{n\to\infty} \frac{\tan n}{n}$ doesn't exist.

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  • $\begingroup$ @AlexM the coefficient is optimal, if you make $\frac{1}{\sqrt{5}}$ smaller, then the theorem fails for the golden ratio $\phi$ or irrational numbers equivalent to $\phi$ (i.e numbers of the form $\frac{p\phi + q}{r\phi + s}$ with $ps-qr = \pm 1$). $\endgroup$ – achille hui Jun 13 '16 at 22:01
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Hint: You need to show that there are infinitely many integer numbers close to multiples of $2\pi$ and infinitely many integer numbers close to $\pi/2+2\pi k$. Numbers close to multiples of $2\pi$ will have a limit of almost $0$ and the others will have a limit almost $\pm$ infinitely large. It is difficult to construct such sequences as $\pi$ has no obvious patterns. But I am sure that someone will give an elegant proof to it.

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    $\begingroup$ $n$ typically denotes positive integers in this type of limits..... $\endgroup$ – N. S. Jun 13 '16 at 20:07
  • $\begingroup$ @N.S. The $\tan$ of positive integers? $\endgroup$ – Noble Mushtak Jun 13 '16 at 20:09
  • $\begingroup$ @NobleMushtak Yes, since $\tan$ is defined at all real numbers, it makes sens for all positive integers... $\endgroup$ – N. S. Jun 13 '16 at 20:12
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    $\begingroup$ Also, $\frac{\tan n}{n}$ tends to $0$ for both of those sequences of values for $n$. So this doesn't help even if the limit was over real values of $n$. $\endgroup$ – JimmyK4542 Jun 13 '16 at 20:12
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    $\begingroup$ Yes it will, because you can always find a sequence of numbers from an interval that takes values between some lower and upper bound and you can find another sequence of numbers that lie in another interval which is not part of the first. Hence, you can construct to limits to this series almost 0 and almost $\pm \infty$. $\endgroup$ – MrYouMath Jun 13 '16 at 20:49

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