$\lim_{a\to\infty} \sin(a)$ if $\lim_{a\to\infty}=0$ I want to find the value of $\lim_{n\to\infty} \sin(\frac{2}{n^3})$, or actually to see if it exists.
Can I show that it is $0$ (or at least exists) with the reasoning that $\lim_{n\to\infty}\frac{2}{n^3}\rightarrow0$ and that $\sin(0)=0$?
Is this reasoning correct, and if so, in the general case or only in my case? 
I.e, is it true that if for a function $f(x)$ where $f(0)=0$ and $\lim_{x\to\infty}x\rightarrow0$, that $\lim_{x\to\infty}f(x)\rightarrow0$?

$$\lim_{n \rightarrow \infty} \sin(\frac{2}{n^3}) = \sin(\lim_{n \rightarrow \infty} \frac{2}{n^3}) = \sin(0) = 0$$
I used the Theorem: If $\lim_{n \rightarrow \infty} a_n = L$ and the function $f$ is continuous at $L$, then $$\lim_{n\rightarrow \infty} f(a_n) = f(L)$$
So, since
$\lim_{n \rightarrow \infty} \frac{2}{n^3} = 0$ and the function $\sin(x)$ is continuous at $0$, then $$\lim_{n\rightarrow \infty} \sin(\frac{2}{n^3})  = \sin(0) = 0.$$
