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Is the following function continous on $\mathbb R$?

$f(x)=\begin{cases}\begin{align} &x\cos \frac 1x, & x\neq 0 \\ &0, & x=0 \end{align}\end{cases}$

I tried to derive it and show the limit from both 0+ and 0- are 0 but I get to infinity plus infinity and such expressions.

This is the derivative: $\frac 1x\sin\left(\dfrac{1}{x}\right)+\cos\left(\dfrac{1}{x}\right)$

Maybe this function isn't continous at all ?

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    $\begingroup$ If you want to check continuity, why do you differentiate? $\endgroup$ – Daniel Fischer Jan 6 '14 at 21:07
  • $\begingroup$ @DanielFischer Hmm if the function has a derivative at a point its continuous. $\endgroup$ – Senishoshitsu Jan 6 '14 at 21:08
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    $\begingroup$ Yes, but a lot of functions are continuous but don't have a derivative. Continuity is a much weaker condition. $\endgroup$ – Daniel Fischer Jan 6 '14 at 21:09
  • $\begingroup$ Ah I see, I made it much more complicated than it should. $\endgroup$ – Senishoshitsu Jan 6 '14 at 21:10
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    $\begingroup$ Bounding the difference $\lvert f(x) - f(a)\rvert$ is the most direct way. That may sometimes be easier to do one side at a time, but not much. Generally, you use the usual theorems (sums, products, compositions etc. of continuous functions are continuous) when you can, and at special points like $0$ here, you have to estimate the difference. I don't see any useful other way than what user did. $\endgroup$ – Daniel Fischer Jan 6 '14 at 21:22
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It is well know, that $$\left|\cos{a}\right|\leq 1$$ for every $a\in\mathbb{R}$

Because of this it holds $$\left|x\cos{1/x}\right|\leq \left|x\right|\cdot\left|\cos{1/x}\right| \leq\left|x\right|\cdot 1 = \left|x\right|$$ Thus $$\lim_{x\rightarrow 0} f(x) = 0 = f(0) $$

which means that $f$ is continous at $x=0$

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  • $\begingroup$ So simple. Thanks. $\endgroup$ – Senishoshitsu Jan 6 '14 at 21:11
  • $\begingroup$ Why the two bars? $\endgroup$ – user63181 Jan 6 '14 at 21:12
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    $\begingroup$ @SamiBenRomdhane just a "TeX"-issue. I made an edit to my answer :) $\endgroup$ – user127.0.0.1 Jan 6 '14 at 21:14
  • $\begingroup$ How do you prove that inequality ? $\endgroup$ – Senishoshitsu Jan 6 '14 at 21:22
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    $\begingroup$ @Senishoshitsu I made an edit. I hope it's more clear now. $\endgroup$ – user127.0.0.1 Jan 6 '14 at 21:30

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