Is $f(x)=x^2\sin\frac{1}{x^2}$ is bounded? Is $f(x)=x^2\sin\frac{1}{x^2}$ bounded?
I have plotted its graph by wolfram Alpha. Here it is.
http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E2sin%281%2Fx%5E2%29+from+x%3D0+to+100
It is easy to see $f$ is bounded by 1. 
How to prove it? Any help would be appreciated.
 A: HINT:
For $x\ne 0$ we have $$ \sin(1/x^2)<1/x^2$$
A: $$\lim_{x\to\infty} x^2 \sin\frac{1}{x^2}$$
$$ = \lim_{x\to\infty} \frac{\sin(x^{-2})}{x^{-2}}$$
$$ = \lim_{x\to\infty} \frac{-2x^{-3}\cos (x^{-2})}{-2x^{-3}}$$
$$ = \lim_{x\to\infty} \cos (x^{-2})$$
$$ = \cos (0)$$
$$ = 1$$
Thus, the function approaches $1$. Further note that if we substitute $x^{-2} = u$ we get $\frac{sin(u)}{u}$. Notice that near $x=0$ the function is nearly $1$, but does not quite reach it... in fact, $$ \lim_{x\to0}\frac{sin(u)}{u} = 1$$
The first two roots $> 0 $ of $\frac{d}{du} \frac{sin(u)}{u} = \frac{u\cos(u)-sin(u)}{u^2}$ are $0$ and 4.49340... and the derivative is negative at $x=1$, so we know that the derivative must be negative in the interval $(0,4.49340...)$. From this, we find that $\frac{sin(u)}{u}$ is bounded by $1$ from $(0,4.49340...)$. The next two critical point is at $7.72525...$, and we find that the derivative is positive in the interval $(4.49340...,7.72525...)$. However, the value of $\frac{sin(u)}{u}$ at this point is $0.12837...$, a great deal less than $1$. Continuing looking at roots of the derivative, we see that $sin(u)$ is periodic from $[-1,1]$, while $\frac{1}{u}$ is constantly decreasing for all $u>0$. Thus, while the function fluctuates between being positive and negative as $u$ increases, each critical point we encounter will have the same $\sin$ value as one of the critical points already discussed; however, $\frac{1}{u}$ will be smaller. As a result, the value at each critical point of $\frac{sin(u)}{u}$ will be closer to $0$ than the value at the previous critical point. Thus, we have shown that the function $\frac{sin(u)}{u}$ is bounded by $1$ for all $u$, and thus so is $x^2 \sin(x^{-2})$.
A: So basically we want to compute $\lim\limits_{x\to\infty}x^2 \sin(\frac{1}{x^2})$. You can make the change of variables $u=\frac{1}{x^2}$. This becomes:
$$\lim\limits_{u\to 0} \frac{\sin(u)}{u}=1$$
A: Result: If a function $f$ is continuous on $\mathbb{R}$ and $\lim_{x\to\pm\infty} f(x)$ exists finitely then $f$ is bounded on $\mathbb{R.}$
Proof: If $\lim_{x\to\infty} f(x)=l,$ then by definition of limit at $\infty,$ for given any $\epsilon>0$ there is some real say $K$ such that $|f(x)-l|<\epsilon$ whenever $x\geq K$ which says that $f$ is bounded on $[K,\infty).$ Similarly we can prove that there exist a real number $k$ such that $f$ is bounded on $(-\infty,k]$  by using $\lim_{x\to-\infty} f(x)=l^{'}$. So $f$ is bounded on $(-\infty,k],$  $[k,K]$(because of continuity) and on $[K,\infty).$ Hence $f$ is bounded on $\mathbb{R}.$ 
Now since $\lim_{x\to\infty} x^{2}sin(\frac{1}{x^{2}})=1,$ and $f$ being continuous(at 0 it is assumed to be 0 value) it is bounded on $\mathbb{R}.$
