The limit was $\lim _{x\to \infty \:}\left(\frac{\tan\left(x\right)}{x}\right)$ I tried several ways to solved it, but I didn't find the answer. I think that it doesn't have a limit, but I want to see others' opinions. What do you think? Thanks in advance.

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    $\begingroup$ What is your reason for beliving it doesn't have a limit? $\endgroup$
    – Spencer
    Commented Jan 17, 2016 at 11:34
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    $\begingroup$ What are the "several ways" that you have tried? $\endgroup$ Commented Jan 17, 2016 at 11:35
  • $\begingroup$ @Spencer Hey Thanks for your comment, well when I tried several techniques That I know, I failed. So I end up drawing the function tanx/x and as you know the curve is repeatable, so we can't guess the limit. That's why :) $\endgroup$ Commented Jan 17, 2016 at 11:36
  • $\begingroup$ @SchrodingersCat I tried to change the variable, then put the tanx as sinx/cosx but no result at the end, so I thought maybe we can solve it by the euler expression and use the commun exp limits we know, but no result too always undefind, and L'hopital rules didn't help too :) $\endgroup$ Commented Jan 17, 2016 at 11:38
  • $\begingroup$ @AmineMarzouki, thats good; now consider $\lim_{x\rightarrow \infty} (sin(x)/x)$, does that exist? Whats the difference from your example? $\endgroup$
    – Spencer
    Commented Jan 17, 2016 at 11:38

2 Answers 2


The limit does not exist. Here is a way to prove it.

You have, for $n \in \mathbb{N}$, $$ \lim _{n\to \infty \:}\left(\frac{\tan\left(2n\pi\right)}{2n\pi}\right)=\lim _{n\to \infty \:}\left(\frac{0}{2n\pi}\right)=0. $$ You also have, for $n \in \mathbb{N}$, $$ \lim _{n\to \infty \:}\left(\frac{\tan\left((2n+1)\frac{\pi}2-\frac1n\right)}{(2n+1)\frac{\pi}2-\frac1n}\right)=\lim _{n\to \infty \:}\left(\frac{\cos(1/n)}{((2n+1)\frac{\pi}2-\frac1n)\sin(1/n)}\right)=\frac1{\pi}\neq 0. $$

  • $\begingroup$ Good Idea thanks for your answer, however I didn't understand the last one, why does $\left(\left(2n+1\right)\frac{\pi }{2}-\frac{1}{n}\right)\sin \left(1/n\right) = \pi $ ? $\endgroup$ Commented Jan 17, 2016 at 12:14
  • $\begingroup$ @AmineMarzouki You may recall that, as $n \to \infty$, we have $\sin(1/n) \sim 1/n$. Thanks! $\endgroup$ Commented Jan 17, 2016 at 12:21
  • $\begingroup$ Still didn't understand it. we have n -> inf so lim as n approach infinity of 1/n is 0 then sin(0)=0. Where did I miss ? $\endgroup$ Commented Jan 17, 2016 at 12:23
  • $\begingroup$ @Armine Marzouki. As $n \to \infty$, you have $$\left(\left(2n+1\right)\frac{\pi }{2}-\frac{1}{n}\right)\sin \left(\frac1n\right) \sim \left(\left(2n+1\right)\frac{\pi }{2}-\frac{1}{n}\right)\frac1n = \frac{2n}{n}\frac{\pi}2+O \left(\frac1n\right) \to \pi.$$ $\endgroup$ Commented Jan 17, 2016 at 12:26
  • $\begingroup$ No that's not what I didn't understand, my problem is why that sin(1/n) = 1/n ? $\endgroup$ Commented Jan 17, 2016 at 12:32

As I see from your comments, you are in the right way. However, you need to prove it mathematically.

HINT: Try to find a sequence $\{a_n\}$ increasing to infinity such that $\{f(a_n)\}$ diverges.

[As you show your work by editing the original post I will keep adding small hints if needed.]

  • $\begingroup$ Thanks for your answer, but could you clarify more your idea? How can we solve a limit with sequences ? can you provide an example ? $\endgroup$ Commented Jan 17, 2016 at 12:05
  • $\begingroup$ My idea is precisely what the user Olivier Oloa wrote in his answer (however, I did not want to do the whole thing since it looks like a homework exercise). Essentially, the idea is to prove that the function is not arbitrarily close to any value whenever $x\to \infty$, which contradicts the definition of limit, and therefore is sufficient to prove its unexistence. Let me show you a simpler example: If $f(x)=1$ in $[2k,2k+1)$ ($k\in \Bbb N$), and $0$ elsewhere, then $f(2k)=1$ and $f(2k+1)=0$ for all $k$, so the function cannot have any limit. $\endgroup$ Commented Jan 17, 2016 at 12:09
  • $\begingroup$ Yep That makes sense now Thank you ! $\endgroup$ Commented Jan 17, 2016 at 12:18
  • $\begingroup$ You're welcome :). $\endgroup$ Commented Jan 17, 2016 at 12:20

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