type of discontinuity $f(x) = \frac{1}{x}\cdot\sin(\frac{1}{x})\cdot\cos(\frac{1}{x})$
$f : \mathbb R \backslash \{0\} \rightarrow \mathbb R$
I need to specify the type of discontinuity at $x_{0} = 0$ (type 1 - jump, type 2 - essential, or removable). Here is what I tried to do:
$f(x) = \frac{1}{x}\cdot\sin(\frac{1}{x})\cdot\cos(\frac{1}{x}) = \frac{1}{2x} \cdot \sin(\frac{2}{x})$Then I tried to calculate the limit when $x \to 0$ but it doesn't seem to have one... Am I even close?

Lisa one can prove discontinuity of this function by sequences. For example,choose $x_{n}=\frac{1}{n{\pi}+\frac{\pi}{4}}$ and $y_{n}=\frac{1}{n{\pi}+\frac{\pi}{2}}$. Then $f(x_{n})=\frac{n\pi}{2}+\frac{\pi}{8}$tends to $\infty$ while $f(y_{n})=0$ and approaches to $0$.
