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I should show that $f':\mathbb{R}\rightarrow\mathbb{R}$ is not continuous at the point $x=0$, where $$ x \mapsto \begin{cases} 2x \cos{\left(\frac{1}{x}\right)}+\sin{\left(\frac{1}{x}\right)} & ,x\neq 0 \\ 0 & ,x=0 \end{cases}$$


My idea is to show that $\lim_{x\rightarrow \infty}f'(x) \neq f'(0)$. I have $f'(0)=0$, and $\lim_{x\rightarrow \infty}(2x \cos(\frac{1}{x})+\sin(\frac{1}{x}))=\infty$. But i am not sure if this is correct. If it is not correct, why not ? And how can i show it than ?

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    $\begingroup$ You have to take the limit at $0$, not at infinity. $\endgroup$ – sxd Apr 2 '13 at 21:45
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Note: on a metric space, a function $g$ is continuous at $x_0$ if and only if $g(x_n)\longrightarrow g(x_0)$ whenever $x_n\longrightarrow x_0$.

Let's try the points $x_n=\frac{2}{(2n+1)\pi}$. Of course, $x_n\longrightarrow 0$. Now $$ f'\left( \frac{2}{(2n+1)\pi}\right)=\frac{4}{(2n+1)\pi}\cos\left( \frac{(2n+1)\pi}{2}\right)+ \sin\left( \frac{(2n+1)\pi}{2}\right)=(-1)^n. $$ Since it admits two subsequences converging to two distinct limits $\pm 1$, $f'(x_n)$ does not converge to any value. In particular, it does not converge to $f'(0)=0$.

So $f'$ is not continuous at $0$.

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  • $\begingroup$ How do you get $x_{n}=\frac{2}{n*\pi}$. Do i have to take the limit at $0$ or $\infty$ ? So now i have $2(\frac{2}{n*\pi})\cos{(\frac{n* \pi}{2})}+\sin{(\frac{n*\pi}{2})}$ $\endgroup$ – Devid Apr 2 '13 at 22:04
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    $\begingroup$ @Devid You need to look at sequences $x_n$ which tend to $0$. I have updated. The best choice is $\frac{2}{(2n+1)\pi}$. It is a good choice because it makes the computation of the cos and the sin easy. $\endgroup$ – Julien Apr 2 '13 at 22:07
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    $\begingroup$ @Devid It is just much easier to do it with sequences than with arbitraray $\epsilon$ and $\delta$ neighborhoods. And this choice of sequence is very natural given the function. $\endgroup$ – Julien Apr 2 '13 at 22:21
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    $\begingroup$ @Devid Try to read this: en.wikipedia.org/wiki/Continuous_function. The keyword is "sequentially continuous", which is the same as "continuous" on a metric space. $\endgroup$ – Julien Apr 2 '13 at 22:26
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    $\begingroup$ @Devid $f'(x)$ is made of two pieces. The lhs term tends to $0$ at $0$. So indeed it boils down to showing that $\sin (1/x)$ does not have a limnit when $x$ tend to $0$. Again, the sequential way is the easier way. As you need to prove it. $\endgroup$ – Julien Apr 2 '13 at 22:58

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