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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

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    $\begingroup$ You are correct. $\endgroup$ – StubbornAtom Jul 21 '16 at 17:08
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    $\begingroup$ 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. $\endgroup$ – DonAntonio Jul 21 '16 at 17:10
  • $\begingroup$ i guess i need to say the function is bounded? is that the correct term? thank you $\endgroup$ – Haonan Chen Jul 21 '16 at 17:11
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    $\begingroup$ Possible duplicate of: math.stackexchange.com/q/1014892/351267. Here's an answer: math.stackexchange.com/a/337998/351267 $\endgroup$ – Zabuza Jul 21 '16 at 22:49
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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.

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  • $\begingroup$ i still haven't learn how to prove or disapprove continuity using epsilon.. so i try to do it with limit $\endgroup$ – Haonan Chen Jul 21 '16 at 17:13
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    $\begingroup$ Have you learned that $f$ is continuous at $a$ if $\lim_{x \to a}f(x)=f(a)$? Here we have $\lim_{x \to a}f(x)\neq f(a)$. $\endgroup$ – Olivier Oloa Jul 21 '16 at 17:15
  • $\begingroup$ yes. i learned that in bc, and that is what i am using to prove its discontinuity in my question. I showed that the limit is bounded instead of 0, therefore the function is not continuous at 0. $\endgroup$ – Haonan Chen Jul 21 '16 at 17:18
  • $\begingroup$ @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. $\endgroup$ – Olivier Oloa Jul 21 '16 at 17:21
  • $\begingroup$ i don't quite get why you want to set 1/x as the latter... I also don't understand how the latter is able to directly show the discontinuity $\endgroup$ – Haonan Chen Jul 22 '16 at 2:24

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