How to reason about the smoothness of a function?

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$Edit$: I use the term "smooth" here to mean "infinitely differentiable".

I would like to ask for some advice on figuring out whether a function is smooth or not, especially when the function is a product, quotient or a composition of other functions. The two example functions which I encountered and which made me ask this are:

$f(x) = \begin{cases}\sin (x)\exp (-\frac{1}{x^2}) & x\neq 0 \cr 0 & x = 0\end{cases}$

and

$g(x) = \begin{cases}\ln (x)\sin (2\pi x) & 0 < x \leq 1 \cr 0 & x = 0\end{cases}$

For the second one it's pretty easy to see it's not differentiable at $0$ just by applying the definition of a derivative, yet if it wasn't defined like that at point zero, I would be lost, as I am with the first one. So yeah, any advice on how to find whether these kinds of functions are smooth or not would be greatly appreciated.

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You can do the first one with this theorem which is a simple consequence of the mean value theorem.

Let $f$ be defined on $(a,b)$ and differentiable except perhaps at the point $c\in(a,b).$ Suppose further that $\lim_{x\to c}f'(x)$ exists. Then $f$ is differentiable at $c$ and $f'(c)= \lim_{x\to c}f'(x)$.

Of course the given $f$ is smooth for $x\neq0$ and when we differentiate it repeatedly, we will obviously get expressions of the form $$f^{(n)}(x)=\exp(-1/x^2)\left(\sin(x)P_n(1/x)+\cos(x)Q_n(1/x)\right)$$ for some rational functions $P_n$ and $Q_n$.

As $x\to0$, $\exp(-1/x^2)->0$ much faster than $P_n(1/x)$ and $Q_n(1/x)$ go to $\pm\infty$ so $$f(n)(0)=0,\ n=0,1,2,\dots$$