# Proving that $\sin(1/x)$ is not continuous at 0

Let $$f(x) = \begin{cases} 0 &\text{ if x=0}\\ \sin(1/x) &\text{ otherwise} \end{cases}$$ Prove that $f$ is discontinuous at $0$

My proof goes like this: for the function to be continuous at 0, the following limit:

$$\lim_{x\rightarrow 0}(\sin(1/x))$$

needs to exist and be equal to 0. Let $$1/x=k$$, I rewrite the limit expression as:

$$\lim_{k\to\infty}(\sin(k))$$

And since this limit oscillates, the limit does not exist. Therefore f(x) is not continuous at 0.

Am I correct? Please tell me if I am doing the right thing! Thank you

• You are correct. – StubbornAtom Jul 21 '16 at 17:08
• The limit doesn't oscillate: the function does as $\;k\to\infty\;$ . A little more formalism may be required, as "oscillating" isn't a mathematical term appliable in this case to prove the limit doesn't exist. – DonAntonio Jul 21 '16 at 17:10
• i guess i need to say the function is bounded? is that the correct term? thank you – Haonan Chen Jul 21 '16 at 17:11
• Possible duplicate of: math.stackexchange.com/q/1014892/351267. Here's an answer: math.stackexchange.com/a/337998/351267 – Zabuza Jul 21 '16 at 22:49

I think you could be more explicit. By writing for example, as $k \to \infty$, $$\frac1x=(4k+1)\frac \pi2 \implies \sin \left( (4k+1)\frac \pi2\right)=1\neq f(0)=0$$ it is clearer.
• Have you learned that $f$ is continuous at $a$ if $\lim_{x \to a}f(x)=f(a)$? Here we have $\lim_{x \to a}f(x)\neq f(a)$. – Olivier Oloa Jul 21 '16 at 17:15
• @HaonanChen Do you get my answer above? Observe it is what you wrote, but instead of taking $1/x=k$ we take $1/x=(4k+1)\pi/2$, the latter is direct to show the discontinuity. – Olivier Oloa Jul 21 '16 at 17:21