Find the range of $f:\ (0,\infty)\ \longrightarrow\ \mathbb{R}:\ x\ \longmapsto\ \sin(\frac{1}{x})$. I'm having trouble with this easy exercise: > Find the range of $f:\ (0,\infty)\ \longrightarrow\ \mathbb{R}:\ x\ \longmapsto\ \sin(\frac{1}{x}).$ I can't analyse the domain of the inverse because $f$ is not invertible, and I don't know how to restrain the domain for it to be. I tried this: If $y$ is such that: $y = \sin(\frac{1}{x})$ $\arcsin(y) = \frac{1}{x}$ $\frac{1}{\arcsin(y)} = x$ Then there are $x$ in this form for $y$. But when $x>0$, $y \in (0,\infty)$, while the image was supposed to be $[-1,1]$. I tried also using continuity, but $(0,\infty)$ is not a compact set... Any help would be appreciated.

View $\sin(\frac{1}{x})$ as a composition of the functions $$g:\ (0,\infty)\ \longrightarrow\ (0,\infty):\ x\ \longmapsto\ \frac{1}{x}\qquad\text{ and }\qquad h:\ \Bbb{R}\ \longrightarrow\ \Bbb{R}:\ x\ \longmapsto\ \sin(x).$$ What is the range of $h(x)$? What does this tell you about the range of $h(g(x))$?