Proof of $\lim_{x \to \infty}\tan x/x$ does not exist There is an answer for this question as follows:
If we approach to infinity with the sequence $a_{n}=n\pi$ then limit is zero, on the other hand if we approach with the sequence $b_{n}=\pi/2+n\pi-1/n^2$ then limit is infinity.
I can not understand the process that is described.
 A: Let's go to definitions. What does it mean for $lim_{x\to\infty}f(x)$ to exist? It means that there exists some $L\in \mathbb{R}\cup\{-\infty, \infty\}$ such that for any sequence $(x_n)$ (where each $x_n$ is in the domain of $f$), that has the property that $lim_{n\to\infty}(x_n)=\infty$, we have that $lim_{n\to\infty}(f(x_n))=L$. This may be slightly hard to parse, so let's put it in words. All this definition is really saying is that regardless of how we approach infinity (this corresponds to the "for any sequence with the property that $lim_{n\to\infty}(x_n)=\infty$" part of the definition) The values of $f$ along this path should approach the same value (this corresponds to the "$lim_{n\to\infty}(f(x_n))=L$ regardless of $(x_n)$" part of the definition). Now let's look at your example of $\tan(x)/x$. If we approach infinity with the sequence $a_n = n\pi$, then we see that the values of $f$ along this path approach zero. Symbolically, we have $\lim_{n\to\infty}f(a_n)=0$. If on the other hand we approach infinity with the sequence $b_n = \pi/2 + n\pi - 1/n^2$, we see $lim_{n\to\infty}f(b_n)=\infty$. Thus the value that $f$ approaches depends on the path we use to approach infinity. The definition tells us that the limit only exists if the value $f$ approaches is independent of the path we choose. So the limit cannot exist.
A: This is a very late response, but something else led me to the function $\frac{\tan{x}}{x}$ and its limit as $x \to \infty$. I came across this post only now.
We can, in fact, prove a more general result for the function $f(x) = \frac{\tan{x}}{x}$ than what is said originally in the post. We can prove the following:
Given any $l \in \mathbb{R}$, there exists a sequence of real numbers, $\langle b_n \rangle_{n \in \mathbb{N}}$, such that $\lim \limits_{n \to \infty} b_n = \infty$ and $\lim \limits_{n \to \infty} f(b_n) = l$.
To see this, let $\epsilon_n = \frac{1}{\alpha n}$ where $n \in \mathbb{N}$ and $\alpha \neq 0$ be any (fixed) real number. Then $\lim \limits_{n \to \infty} \epsilon_n = 0$. Now define $b_n = \frac{\pi}{2} + n \pi - \epsilon_n$ for $n \in \mathbb{N}$. Then $\lim \limits_{n \to \infty} b_n = \infty$, and you can prove that $\lim \limits_{n \to \infty} f(b_n) = \frac{\alpha}{\pi}$. Given any $l \in \mathbb{R}$, $l \neq 0$, if you now choose $\alpha = \pi l$, then it is clear that $\lim \limits_{n \to \infty} f(b_n) = l$.
When $l = 0$, let $b_n = n \pi$. Then $\lim \limits_{n \to \infty} f(b_n) = 0$.
Also, if you let $\epsilon_n = \pm \frac{1}{n^p}$ where $p > 1$, you can prove that $\lim \limits_{n \to \infty} f(b_n) = \pm \infty$.
What this argument shows is that you can "adjust" the limit of $f(b_n)$ to anything you want. Thus there is no unique limit.
