<s>[INST] To check whether a function is unbounded 
Show that $f:[-1,1]\to \mathbb{R}$ defined by
  $f(x)=2x\sin\frac{1}{x^2}-\frac{2}{x}\cos\frac{1}{x^2}, ~~x\neq 0$ and $f(0)=0$ is unbounded on every neighbourhood of $0$.

Attempt:
$|f(x)|=|2x\sin\frac{1}{x^2}-\frac{2}{x}\cos\frac{1}{x^2}|\leq |2x\sin\frac{1}{x^2}|+|\frac{2}{x}\cos\frac{1}{x^2}|\leq 2+|\frac{2}{x}|$. At the neighbourhood 
 of $x=0$, $\frac{1}{x}$ is unbounded, so $f(x)$ cannot be written as $|f(x)|\leq K$ for some positive $K$. Thus $f(x)$ is unbounded. 
Is this proof right?
 [/INST] Let $x_k = \frac{1}{\sqrt{2k\pi}}$ where $k\in \mathbb{Z}, k > 0$.
Then $$f \left( \frac{1}{\sqrt{2k\pi}} \right)=-2\sqrt{2k\pi}$$
As $k\to \infty, x_k \to 0, |f(x_k)| = 2\sqrt{2k\pi}\to \infty$.
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