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?
 A: 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 $0$.
A: What will happen with $f(x)$ near $0$ is that it will oscillate ever faster and wider. It will take on all real values arbitrarily near $0$. The reason you can't find the limit is because it doesn't exist.
To be more detailed you can consider $x_n = 4/n\pi$ and calculate the value of $f(x_n)$:
$$f(x_n) = {1\over2x_n}\sin{2\over x_n} = {n\pi\over 8}\sin{n\pi\over2}$$
This means for example that $f(x_{2n})=0$, $f(x_{4n+1}={(4n+1)\pi\over8}$ and $f(x_{4n+3})={(4n+3)\pi\over8}$. These sequences goes to $0$, $+\infty$ and $-\infty$ while the argument goes to zero. Such behavior is not consistent with $\lim_{x\to0}f(x)$ existing. 
How you classify your discontinuities is a bit unclear, but I'd assume that it's what you call an essential discontinuity.
