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I am trying to understand Daniel Fischer's answer to this question Antiderivative of $\sin(\frac{1}{x})$

I understand the ideas and every step except for the part where he said

The first part $x^2\cos(x^{−1})$ is easily seen to be differentiable at $0$ with derivative $0$ there

But isn't $x^2\cos(x^{−1})$ not defined at $x=0$? So how can we talk about its derivative there? I can see that the limit at $0$ exists, but it's still not defined there. Is he using some properties of $F'$ that I am not seeing? Or do we actually have somehow have this point defined to be zero at zero? Since by Show that the function $g(x) = x^2 \sin(\frac{1}{x}) ,(g(0) = 0)$ is everywhere differentiable and that $g′(0) = 0$ we know that the derivative is zero if we define $x^2\cos(x^{−1})$ to take the value zero at zero, so do we have this? Is there something about how we define $F$ and/or $F'$ that forces this to be true?

I can't really be sure of this, any help is appreciated.

Mathemphetamine
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  • I'd say with about a 99.9% chance of probability that yes, they are are talking about the function with the "hole" removed, i.e. by defining the function at 0 to be its limit. – Alan Mar 02 '21 at 16:53

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$x^2 cos(1/x)$ is actually defined at $x=0$. It can be seen by applying squeeze theorem. As $-1\leq cos(1/x) \leq 1$ for all $x$ hence $-x^2\leq x^2 cos(1/x)\leq x^2$. The limits $-x^2$ and $x^2$ as $x$ approaches $0$ are $0$. Thus at $x=0$, $x^2 cos(1/x)$ is squeezed between $0$'s and hence itself $0$.

HARVEER RAWAT
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