Value of $x^2 \sin (\frac1x)$ at $x=0$ What is the value of $y=x^2 \sin(1/x)$ at $x=0$?  

I see that $x^2 =0$ but $\sin (1/x)$ is undefined.
More generally: if a function made up of a product of functions,  like $y= a(x)b(x)c(x)d(x)\dots$, then at a specific value of $x$ if one of the functions is $0$ and all the other functions are undefined, does it still mean that $y=0$?

The function $x \, \longmapsto \, x^{2} \sin \Big( \frac{1}{x} \Big)$ is not defined at $x = 0$ because "$1/0$" does not exist. Note also that this function is continuous on $\mathbb{R}^{\ast}$. 
However, this is not all we can say. Since the function $x \, \longmapsto \, \sin \Big( \frac{1}{x} \Big)$ is bounded, one can prove that :
$$ \lim \limits_{x \to 0} x^{2} \sin \Big( \frac{1}{x} \Big) = 0.$$
As a consequence, even though the function $x \, \longmapsto \, x^{2} \sin \Big( \frac{1}{x} \Big)$ is not defined at $x=0$, it has a finite limit as $x$ goes to $0$. So, this function can be extended to a new function, say $\overline{f}$, defined on $\mathbb{R}$ the following way :
$$ \overline{f}(x) = 
\begin{cases} x^{2} \sin \Big( \frac{1}{x} \Big) & \text{if } x \neq 0 \\[2mm]
0 & \text{if } x = 0
\end{cases}
$$
This new function $\overline{f}$ is defined at $x=0$ and continuous on $\mathbb{R}$. In a very unformal, non-mathematical way, we could say that the value of the function $x \, \longmapsto \, x^{2} \sin \Big( \frac{1}{x} \Big)$ at $x=0$ is $0$. But this is not really the truth !
