Show $f(x)=2x \sin{\frac{1}{x} -\cos{\frac{1}{x}}}$ is discontinuous at $0$

Question

Show $f$ is disconinuous at $0$ where $f$ is:

$ f(x)= \begin{cases} 2x \sin{\frac{1}{x} - > \cos{\frac{1}{x}}}&\text{if}\, x \neq 0\\ 0&\text{if}\, x=0 > \end{cases} $

My attempt (proof by contradiction):

Suppose $f$ is continuous at $0$ then $\forall \epsilon \gt 0$, $\exists \delta \gt 0$ s.t. $\left| f(x)-f(0) \right| \lt \epsilon$ whenever $|x-0| \lt \delta$.

Choose $\epsilon = \frac{1}{2}$, then by hypothesis $\exists \delta \gt 0$ s.t. the above condition is true.

Now pick $x = \frac{1}{n \pi}$ where $n \in \mathbb{N}$ is some natural number. Then $x \gt 0$.

For a sufficiently large $n$, $|x-0| = x \lt \delta$.

But $f\left(\frac{1}{n \pi}\right)$ $=\frac{2}{n \pi} \sin{(n \pi)} - \cos{(n \pi)}$ $=0 - (-1)^n = (-1)^{n+1}$

So $|f(x) - f(0)| = |(-1)^{n+1}-0|=1 \gt \frac{1}{2}$ which is a contradiction. Hence, $f$ must be discontinuous at $0$.

$\Box$

I would like some alternative proofs, as well as comments or suggestions to my proof.

Answer 1

Hint: Continuity can be proved using limits as well. For instance, $\sin(\frac{1}{x})$ is bounded, so what can be said about $x\sin(\frac{1}{x})$ when $x \to 0$? What about the second term, $\cos(\frac{1}{x})$?

Answer 2

We have $f(0)=0$. So to show that $f$ is not continuous at $x=0$, it is enough to show that it is not true that $\lim_{x\to 0} f(x)= 0$.

Suppose to the contrary that the limit exists and is equal to $0$. Then for any $\epsilon\gt 0$, there is a $\delta\gt 0$ such that if $|x|\lt\delta$, then $|f(x)|\lt\epsilon$.

Pick $\epsilon=\frac{1}{2}$. We show there is no $\delta$ with the required property.

If $x\lt 0$, then $f(x)\lt 0$. In particular, if $x\lt 0$, then $|f(x)|\gt 0$. It follows that there is no $\delta$ such that $|x|\lt \delta$ guarantees that $|f(x)|\lt \epsilon$.