is $x\sin(1/x)$ bounded? and how can I prove the derivative of it is not bounded? I thought that $x\sin(1/x)$ is not bounded, that it's only bounded below.
Am I correct?(there's no condition nor variation in this question)
and also how can I prove that the derivative of $\sin(1/x)$, which is $\sin(1/x)-(\cos(1/x)/x)$, is not bounded?
Thankyou
 A: It's not clear what you're asking about, as you have different functions in the title and in the question. But both of them are bounded on their entire domains, which is all reals except $0$ (i.e. $(-\infty,0)\cup(0,+\infty)$).
The function $f(x)=\sin(1/x)$ is bounded simply because $-1\le\sin\theta\le1$ for any real $\theta$.
The function $g(x)=x\sin(1/x)$ is also bounded, for the following two reasons: $\lim\limits_{x\to0}\left[x\sin(1/x)\right]=0$ by the Pinching (Squeeze) Theorem and $\lim\limits_{x\to\infty}\left[x\sin(1/x)\right]=1$ because, substituting $t=1/x$,
$$\lim_{x\to\infty}\left[x\sin(1/x)\right]=\lim_{x\to\infty}\left[\frac{\sin(1/x)}{1/x}\right]=\lim_{t\to0}\left[\frac{\sin(t)}{t}\right]=1.$$
You're also confused with your derivatives between these two functions. What you stated to be the derivative of the first function is in fact the derivative of the second one:
$$f'(x)=\left[\sin(1/x)\right]'=-\frac{\cos(1/x)}{x^2}$$
and
$$g'(x)=\left[x\sin(1/x)\right]'=\sin(1/x)-\frac{\cos(1/x)}{x}.$$
The latter one is not bounded because of the $\frac{\cos(1/x)}{x}$ term in it. As $x\to0$, cosine is bounded, so dividing by smaller and smaller values of $x$ causes some of its values to be arbitrarily large by magnitude. Just to make it clear: it does NOT go to any limit, not even to infinity as a limit, because it oscillates. But it at some points it does take on arbitrarily large values (by absolute value).
For example: for $x=\frac{1}{2\pi n}$, where $n$ is an integer, we get
$$g'(x)=g'(1/(2\pi n))=\sin(2\pi n)-\frac{\cos(2\pi n)}{1/(2\pi n)}=0-\frac{1}{1/(2\pi n)}=-2\pi n.$$
