Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm doing an exercise that asks me to prove that $f$ is continuous using a $\epsilon$-$\delta$ proof. I have that $$ f(x) = \begin{cases} x\cdot \sin \frac1x,&x\neq 0 \\ 0,&x = 0 \end{cases} $$ I've already managed to show this property for $x=0$. How can I show it for $x \ne 0$, also using a $\epsilon$-$\delta$ proof?

Thank you very much.

share|cite|improve this question
Do you want to prove the case $x\neq 0$ also using $\epsilon$-$\delta$? – Thomas Jan 9 '13 at 17:04
@Thomas Yes :-) – Gabriel Bianconi Jan 9 '13 at 17:07
I know, this question is answered. But would it have been okay to use a series argument and then say: Well, there is a theorem which tells us $\varepsilon-\delta$-continuity is the same as sequence-continuity. So there has to be a $\delta \gt 0$ fulfilling your requirements. – Keba Jan 22 '14 at 20:34
up vote 4 down vote accepted

Let $a\neq 0$. By the triangle inequality, $$\begin{array}{ccc} \left|f(x)-f(a)\right| &=& \left|x\sin \frac1x-a\sin \frac1a\right| \\ &=& \left|x\sin \frac 1x-a\sin \frac 1x+a\sin \frac 1x-a\sin \frac1a\right| \\ &\le& \left|x-a\right|\left|\sin \frac 1x\right|+a\left|\sin \frac 1x-\sin \frac1a\right| \\ &<& \delta+a\left|\sin \frac 1x-\sin \frac1a\right| \end{array}$$ It all comes down to bounding the second term. By the trigonometric identity \begin{equation}\sin \alpha-\sin \beta=2\sin \frac{\alpha-\beta}2\cos \frac{\alpha+\beta}2\end{equation} we have $$\begin{array}{ccc} \left|\sin \frac 1x-\sin \frac1a\right| &=& \left|2\sin \frac{\frac1x-\frac1a}{2}\cos\frac{\frac1x+\frac1a}{2} \right| \\ &=& 2\left|\sin \frac{x-a}{2xa}\cos\frac{x+a}{2xa} \right| \\ &\le& 2\left|\sin \frac{x-a}{2xa}\right|\end{array}$$

Because $\left|\sin \alpha\right|\le \alpha$, \begin{equation}2\left|\sin \frac{x-a}{2xa}\right|\le 2\left|\frac{x-a}{2xa}\right|=\frac{\left|x-a\right|}{\left|x\right|\left|a\right|}\end{equation} As $\left|x-a\right|<\delta\implies \left|x\right|>\left|a\right|-\delta$, the situation is simplified if we choose $\delta<\frac{\left|a\right|}2$. Then, \begin{equation}\left|x-a\right|<\delta\implies \left|x\right|>\left|a\right|-\delta>\frac{\left|a\right|}2\implies \frac1{\left|x\right|}<\frac{2}{\left|a\right|}\end{equation} and so \begin{equation}\left|\sin \frac 1x-\sin \frac1a\right|\le\frac{\left|x-a\right|}{\left|x\right|\left|a\right|}<\frac{2\delta}{a^2}\end{equation} I belive you can finish this off.

share|cite|improve this answer
Thank you very much! – Gabriel Bianconi Jan 9 '13 at 17:15
@Nameless I've made a little edit. The LaTeX was rolling off the screen. I hope you don't mind. Just check I haven't made an error. – Fly by Night Jan 9 '13 at 17:46
@FlybyNight It's much better now. I don't see any error. Thank you very much – Nameless Jan 9 '13 at 17:50
@Nameless: I have observed that you are very good at $\epsilon-\delta$ proofs :-) – Parth Kohli Jan 9 '13 at 17:55

The function $f$ is continuous on $\Bbb R$ if and only if it is continuous at any point of $\Bbb R$. Since $$ f(x) = a(x)b(x) $$ for $x\neq 0$ and functions $a,b$ are continuous for $x\neq 0$, their product $f$ is also continuous for any $x\neq 0$.

share|cite|improve this answer
Thank you! But is it possible to show this with an epsilon-delta proof? – Gabriel Bianconi Jan 9 '13 at 17:04
The exercise list asks me to "Prove using epsilon-delta arguments." and then lists several functions. I believe they should be used in every step. I have a separate proof for the theorem of the product, but I don't think it can be used for my purpose (if there were no restrictions, it would clearly be a better option). – Gabriel Bianconi Jan 9 '13 at 17:10
@GabrielBianconi: yeah, in that way the solution by Nameless seems to fit better. – Ilya Jan 9 '13 at 17:12
I downvoted you (sorry). The OP wanted an $\epsilon-\delta$ proof. – Thomas Jan 9 '13 at 17:17
@Thomas: thanks for responding (+1)! As I explained above, I do agree that the proof by Nameless is better in such case, and I didn't expect OP is required to give $\varepsilon$-$\delta$ proof for the case when there is much easier proof available - as you can see Nameless essentially first decomposed the difference a-la the proof of continuity of the product of two continuous functions. – Ilya Jan 9 '13 at 17:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.