Find the range of $f:\ (0,\infty)\ \longrightarrow\ \mathbb{R}:\ x\ \longmapsto\ \sin(\frac{1}{x})$.

Question

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.

Answer 1

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 $h(g(x))$?

Answer 2

you know that the range must lie within $[-1,1] = \sin([2\pi,4\pi]) $, right ?

and you have for $g(x) = \frac{1}{x}$

$$g([\frac{1}{4\pi},\frac{1}{2\pi}] \subset (0,\infty)) = [2\pi,4\pi] $$

I'm sure you know how to conclude.