type of discontinuity

Question

$f(x) = \frac{1}{x}\cdot\sin(\frac{1}{x})\cdot\cos(\frac{1}{x})$
$f : \mathbb R \backslash \{0\} \rightarrow \mathbb R$

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?

If the limit doesn't exist, which types of singularities does that rule out? — Daniel Fischer, Dec 17 '15 at 10:42
Have you ever drawn $\sin(1/x)$ before? What do you think happens when you multiply it with $\frac1x$? It's not a real proof, but it should tell you exactly what kind of discontinuity this is. — Arthur, Dec 17 '15 at 10:52
But how do I prove the limit doesn't exist? I can't use lhopotal's rule — Lisa, Dec 17 '15 at 11:06

score 1 · Answer 1 · edited Dec 17 '15 at 12:43

1

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 $0$.

edited Dec 17 '15 at 12:43

skyking

16,654

answered Dec 17 '15 at 12:07

M.R. Yegan

672

skyking · Answer 2 · 2015-12-17T12:43:01.823

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.

type of discontinuity

2 Answers2