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I saw a question asking for the limit

$$\lim_{n \to \infty}\frac{\tan(n)}{n}.$$

At first I thought that the limit assumed $n$ to be a real number. So I gave the advice to use $\pi/2+2\pi k$ and $2\pi k$ as two sequences with different limits. The real limit for $x\to \infty$, in which $x \in \mathbb{R}$, is much easier to handle than the limit $n \to \infty$, in which $n \in \mathbb{N}$.

Here is my question:

Is there a mathematical theorem that is linking the integer limit $$\lim_{n\to \infty}f(n)$$ to the real limit $$\lim_{x\to\infty}f(x)?$$

Is the equidistribution theorem such a mathematical statement?

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    $\begingroup$ If the latter exists then it is the former. The former could exist by itself, in which case knowing that the latter doesn't exist doesn't really tell you anything. $\endgroup$
    – Ian
    Jun 13, 2016 at 21:17
  • $\begingroup$ @Ian is using the fact that for a convergent sequence, any subsequence must also converge to the same limit. (here were assuming were in the extended reals, so we can say the sequence converges to infinity, instead of saying it diverges). So if these two subsequences have different limits, what does that tell you about the convergence of the entire sequence? $\endgroup$
    – SquirtleSquad
    Jun 13, 2016 at 21:20
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    $\begingroup$ If $f$ is monotonic, then if the natural number limit exists, then the real limit exists. Also, if both exist, they are equal, of course. $\endgroup$
    – Thomas Andrews
    Jun 13, 2016 at 21:21
  • $\begingroup$ Be careful, @Merlin, some people use "diverges" to just mean "does not converge to a real," others use it to mean converges to $+\infty$ or $-\infty$ or (sometimes) to the one-point $\infty$. And Ian's statement is not about sequences, since the real case is not a sequence in the usual sense of the term. $\endgroup$
    – Thomas Andrews
    Jun 13, 2016 at 21:23
  • $\begingroup$ @Merlin I don't think you argument is adequate to this case: $\frac{\tan x}{x}$ does not have a limit as $x\to\infty$ basically because you can always find, say, a sequence of real numbers such that $\tan x_n\ge n x_n$ and $x_n\to\infty$ (because you can freely approach a vertical asymptote while changing $x$ of at most $\pi$). However, proving (or disproving) that this can be done by choosing $x_n$ to be natural numbers is quite more sketchy, since in order to make $\tan k$ grow to infinity, you must make $k$ grow to infinity as well. And if you can't do something of sort, then $\lim=0$ $\endgroup$
    – user228113
    Jun 13, 2016 at 21:28

2 Answers 2

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The equidistribution theorem is a useful tool for proving that some limits over $\mathbb{N}$ (do not) exist or some series are (not) converging, but in general $\lim_{n\to +\infty}f(n)$ may exist even if $\lim_{x\to +\infty}f(x)$ does not: just take $f(x)=\sin(\pi x)$. On the other hand, it is trivial that $\lim_{x\to +\infty}f(x)=C$ implies $\lim_{x\to +\infty}f(n)=C$. In our case $\lim_{x\to +\infty}f(x)$ does not exist and neither it does $\lim_{n\to +\infty} f(n)$, but that is tricky. Take a convergent $\frac{p_k}{q_k}$ of the continued fraction of $\frac{\pi}{2}$ with $q_k$ being odd: we have $$ \left| \frac{p_k}{q_k}-\frac{\pi}{2} \right|\leq \frac{1}{q_k^2}, $$ hence $|\sin(p_k)|$ is less than $\frac{1}{q_k}$ apart from $1$ and $\cos(p_k)$ is less that $\frac{1}{q_k}$ apart from zero, hence if $q_k$ is big enough we have $\left|\frac{\tan(p_k)}{p_k}\right|\geq\frac{1}{2}$, so $\limsup_{n\to +\infty}|f(n)|\geq\frac{1}{2}$. With the same technique we may also find a subsequence proving $\liminf_{n\to +\infty} |f(n)|\leq \frac{1}{3}$, so the limit does not exist.

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  • $\begingroup$ Now, can we apply the equidistribution theorem here to establish the link between the limit in $n$ and the limit in $x$. $\endgroup$
    – MrYouMath
    Jun 14, 2016 at 21:28
  • $\begingroup$ @MrYouMath: a posteriori, for sure. $\endgroup$
    – Jack D'Aurizio
    Jun 14, 2016 at 21:32
  • $\begingroup$ What do you mean by that? $\endgroup$
    – MrYouMath
    Jun 14, 2016 at 21:34
  • $\begingroup$ @MrYouMath: well, if we use the properties of some continued fraction to prove that a limit over $\mathbb{N}$ does not exist, the limit over $\mathbb{R}^+$ does not exist either, but the last part is trivial, so we are nuking a mosquito :D $\endgroup$
    – Jack D'Aurizio
    Jun 14, 2016 at 21:36
  • $\begingroup$ No, I want a statement which goes from real limit to natural limit. Not the other way round :D. $\endgroup$
    – MrYouMath
    Jun 14, 2016 at 21:38
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If a function $f$ is continuous, then for any sequence $x_n\to x_0$, $$\lim_{x_n\to x_0} f(x_n)=f(x_0),$$ and in particular if $x_0=\infty$, $$\lim_{n\to\infty}f(n)=\lim_{x\to\infty}f(x)$$ for real $x$ and natural $n$. However this is not the case since $\tan$ is not continuous, and in fact it does not even make sense to say $$\lim_{x\to\infty}\frac{\tan x}{x}$$ because this is only definied if all sequences converging to infinity give the same limit.

So in short, yes, you have to use natural numbers, which does make the question more difficult but also more interesting.

In fact I am starting to think that the limit does not exist in this case either, since the decimal expansion of $\pi/2$ has arbitrarily long sequences of zeros (I don't know if this is even known?), and so $\tan n$ becomes arbitrarily large. But I may be wrong.

EDIT: I'm sorry I just realized Jack D'Aurizio answered that beautifully.

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