How to show whether this limit exists or not? It's my intuition that
$$\lim_{x\to+\infty} \frac{\sin(x+\frac1x)}{\sin(x)}$$
does not exist.  And I have been working on proving it. I have tried Heine's Theorem but for this moment I get stuck and don't know how to find a counter-example sequence. If that doesn't work, I must turn to Cauchy's criterion, which I think will be the last resort. 
Anyhow, please help me prove , or disprove my intuition.  Best regards!
 A: It is indeed the case that the limit does not exist.  Denoting
$$
f(x) = \frac{\sin(x+\frac1x)}{\sin(x)}
$$
we have an infinite discontinuity at $x = \pi n$ for any integer $n$, with $\lim_{x \to n\pi^-}f(x) = -\lim_{x \to n\pi^+}f(x)$.
This alone is sufficient to conclude that the limit does not exist.
You should be able to find a counterexample sequence of the form $2 \pi n - \epsilon_n$ where $\epsilon_n > 0$ is chosen to be sufficiently small (and possibly dependent on $n$).
A: Consider the following decomposition:
$$\frac{\sin(x + 1/x)}{\sin x} = \cos(1/x) + \cot x\sin(1/x)$$
Since the first term goes to $1$ as $x\to\infty$, it is sufficient to prove that $\cot x\sin(1/x)$ does not have a limit. But $\cot x$ has poles at $x=n\pi$, and $\sin(1/x)$ is a well-defined nonzero continuous function on $x\ge1$, so $\cot x\sin(1/x)$ takes on arbitrarily large positive and negative values near each $x=n\pi$ and hence has no limit as $x\to\infty$.
To be more precise, let $x=n\pi+\delta$, where $\delta$ is a suitably small number to be chosen shortly. Then since $\frac d{dx}\tan x=1$ at $x=n\pi$, $|\cot x|\ge \frac 1{2|\delta|}$ for small enough $\delta$ (you can find the bound explicitly without trouble) with $\cot x$ large positive when $\delta>0$ and large negative when $\delta<0$, and then we have a means of generating points far away from any purported limit. If $\cot x\sin(1/x)\to\ell$, then supposing that $|\cot x\sin(1/x)-\ell|\le1$ for $x\ge N$, pick any $n$ such that $n\pi\ge N$ and then for $x=n\pi+\delta$, where $\delta$ is chosen so that $|\cot x|\ge \frac 1{2|\delta|}$ and $\delta\le\frac{\sin(1/((n+1)\pi))}{2(|\ell|+2)}$, we have $|\cot x\sin(1/x)|\ge|\cot x|\sin(\frac1{(n+1)\pi})\ge (|\ell|+2)$ (since $\sin(1/x)$ is a decreasing function for $x\ge\frac2\pi$), a contradiction.
