# How to evaluate $\lim\limits_{n\rightarrow \infty} \sin (1/n)$?

I know that the $\lim\limits_{n\rightarrow \infty} \sin \left( \frac{1}{n} \right)= 0$

But I am not sure of the right workings. My attempt:

1. As $n$ tends to infinity, $\frac{1}{n}$ will tend to $0$.

2. Therefore, $\sin \left(\frac{1}{n}\right)$ will tend to $0$ as $x \rightarrow \infty$ since $\sin(0)=0$

Is my reasoning correct? I am especially doubtful of point number 2 as I am not aware of any rules that allow that kind of operation. Are there more complete/rigorous ways to answer this questions?

Your reasoning is correct, since it is a property of limits that $$\lim_{n \to a} f(g(x)) = f \left( \lim_{n \to a} g(x) \right),$$ even when $a = \pm \infty$, as long as $f$ and $g$ are continuous.
• I am aware of that property but not for the case when $a=\pm \infty$ (My text did not mention it). Could you provide some links that discuss this aspect of the property? I only managed to find discussions assuming $c$ is a point... – mauna Dec 11 '14 at 18:20
You are right, and the "rule", which allows this "kind of operation" is called continuity. In detail: Since $\sin$ is continuous at $x=0$, we can move the limit into the function, so $\lim_{x\to\infty}\sin(1/x)=\lim_{y\to0}\sin(y)=\sin\left(\lim_{y\to0}y\right)=\sin(0)=0$.