# $\epsilon$-$\delta$ proof that $f(x) = x \sin(1/x)$, $x \ne 0$, is continuous

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.

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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

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 $$\sin \alpha-\sin \beta=2\sin \frac{\alpha-\beta}2\cos \frac{\alpha+\beta}2$$ 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 $\left|\sin \alpha\right|\le \alpha$, $$2\left|\sin \frac{x-a}{2xa}\right|\le 2\left|\frac{x-a}{2xa}\right|=\frac{\left|x-a\right|}{\left|x\right|\left|a\right|}$$ 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, $$\left|x-a\right|<\delta\implies \left|x\right|>\left|a\right|-\delta>\frac{\left|a\right|}2\implies \frac1{\left|x\right|}<\frac{2}{\left|a\right|}$$ and so $$\left|\sin \frac 1x-\sin \frac1a\right|\le\frac{\left|x-a\right|}{\left|x\right|\left|a\right|}<\frac{2\delta}{a^2}$$ I belive you can finish this off.

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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 $x\neq 0$.

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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