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The function is $2x \sin(\frac{1}{x}) - \cos(\frac{1}{x})$. I have to show that this function is not continuos at $x = 0$. But I get the result that it is continuous. Am I wrong? Can you provide a good answer for this one?

F(x) = x^2 sin(1/x) when x is not equal to 0. F(x) = 0 when x = 0. I have to show that f(x) is differentiable everywhere, but df/dx is not continuos at x = 0.

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    It is not even defined at $x=0$, how can it possibly be continuous? – 2017-02-07
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    The function is not defined at $x=0$ currently. Do you have it defined separately at $0$? You cannot comment on continuity at a point where it is not defined. – 2017-02-07
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    Then how do we show that result? I can't just say that it is not defined at x = 0. So what can I write as the answer? – 2017-02-07
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    "I can't just say that it is not defined at x = 0". Yes you can. That's perfectly legitimate. – 2017-02-07
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    I would guess the question should *really* be to show that the given function does not have a continuous extension which includes $0$ in its domain. – 2017-02-07
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    Is is $2x\sin(\frac 1x) - \cos(\frac 1x)$ or is it $2x(sin(\frac 1x) - \cos(\frac 1x))$. If I'm not mistaken the latter has a limit of 0 as x tends to 0 whereas the first has no limit at all. The latter can be extended to a continuous function by defining $h(x) = 0$. The first can not. But the second is not continuous at x = 0 because it is undefined at x = 0. It *IS* that simple. – 2017-02-07
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    What is the actual question? In full. – 2017-02-07
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    I have posted the question. Can you give me a seperate answer please? – 2017-02-07
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    Okay, so what is $F'(0)$? – 2017-02-07
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    It can get from f(x) – 2017-02-07
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    @fleablood what should I write as the answer? Can you give me a seperate answer? – 2017-02-07

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"$F(x) = x^2 sin(1/x)$ when $x$ is not equal to 0. $F(x) = 0$ when $x = 0$. I have to show that $f(x)$ is differentiable everywhere, but $df/dx$ is not continuos at $x = 0$".

Let $h(x) = x^2 \sin(\frac 1x)$.

$F(x) = h(x); x\ne 0; F(x) = 0; x = 0$.

If $x \ne 0$ then $F$ is differentiable and continuous if $h$ is.

$h$ is. I'll leave that to you. ($h'(x) = 2x\sin(\frac 1x) - \cos (\frac 1x)$. Also left to you.)

At $x=0$, $F$ is continuous if $\lim_{x\rightarrow 0}h(x) = F(0) = 0$.

It does. I'll leave that to you. (Hint: Squeeze Theorem.)

At $x=0$, $F'$ is differentiable if $\lim_{k\rightarrow 0}\frac {F(0+k) - F(0)}k=\lim_{k\rightarrow 0}\frac {h(k)}k$ exists.

It does. And it is equal to $0$. I'll leave that to you. (Hint: Again with the Squeeze Theorem.)

So $F$ is continuous and differentiable everywhere.

$F'(x) = h'(x); x \ne 0$ and $F'(0) = \lim_{k\rightarrow 0}\frac {F(0+k) - F(0)}k = 0$.

$F'$ is continuous at $x = 0$ only if $\lim_{x\rightarrow 0} F'(x) = \lim_{x\rightarrow} h'(x)= F'(0) = 0$.

It doesn't. I'll leave that to you. (Hint: $\lim_{x \rightarrow \infty} \cos x = ???$)