what is the value when we multiply zero and infinity? We know that $$\lim_{x \to 0} x \sin \left(\frac{1}{x} \right) = 0$$ 
When $x \to 0$ then $\sin(1/x)$ is undefined and multiplying this by $0$, the ultimate result is $0$. 
Why is this not the case when we multiply $0$ and $\infty$?
 A: The function $f(x)=\sin\left(\frac{1}{x}\right)$ only takes the values $f \in [-1,1]$.
Thus, you are multiplying $0$ by a finite value, and therefore, the limit converges towards zero.
However, since $\infty$ is not finite like $\sin\left(\frac{1}{x}\right)$ is, when multiplied by $0$, the result is undefined.
A: In the extended reals, $0 \cdot \infty$ is undefined.
We define arithmetic to leave it undefined for many of the same reasons why $0/0$ is left undefined.
In particular, we cannot continuously extend multiplication to this case. To extend multiplication continuously, the limit $\lim_{(x,y) \to (0,\infty)} xy$ must be defined; i.e. we must get the same result no matter what path we take approaching $(0, \infty)$.
Here's one path that converges to $1$:
$$ \lim_{x \to 0} x \cdot \frac{1}{|x|} $$
Here's another path that converges to $3$:
$$ \lim_{x \to 1} |x^3-1| \cdot |\cot(x-1)| $$
Since we got two paths with different values, the limit doesn't exist; it is impossible to define multiplication in a way that it is continuous at $(0,  \infty)$.
