Prove that $f:[0,\infty)\to\mathbb{R}$ where $f(x) := {1\over x}\cos({1\over x}),x>0$ ,does $f$ has the intermediate value property on $[0,\infty)$? Prove that $f:[0,\infty)\to\mathbb{R}$ where $f(x) := {1\over x}\cos({1\over x}),x>0$ does $f$ has the intermediate value property on $[0,\infty)$?
Attempts: In $\mathbb{R}$, if $f$ is continuous then it has intermediate value property and as we know ${1\over x}\cos({1\over x})$ are continuous  on  $\mathbb{R}$ so it has the intermediate value property but it doesn't seem to be that straight forward.

This answer is motivated by a question in the comments from the OP @Mathematics.
Note that $\cos(2\pi n)=1$ and that $\cos(2\pi n + \pi) = -1.$ It follows that the function $f(x)=(1/x) \cos(1/x)$ (with $x>0$, we don't need a definition of $f(0)$ for this approach) has the following particular values:
$$f(1/(2 \pi n))=2 \pi n,$$
$$f(1/(2 \pi n + \pi))=-(2 \pi n + \pi).$$
So given any real number $c$ we can choose $n$ large enough so that $c$ lies in the open interval $I_n=(-(2\pi n + \pi), 2 \pi n).$ By the intermediate value theorem $f(x)=c$ for some $c \in I_n.$ So we can say that $f(x)$ takes on every real number value near (but no need for at) $0$. And looking at the above sequence of intervals $I_n$ we can in fact say that $f(x)$ takes on every real number infinitely often.
