Is it a valid $\lim_{x\to 2} (0\cdot x)$? Just curious.
Is it a constant limit i.e $\lim_{x\to 2} (0\cdot x)=0$, since all values will equal to $0$ and tends to nothing. And it doesn't involve infinity, so cannot be indeterminate.
 A: Notwithstanding the standard terminology where “tends to” is used, you should think to this only in terms of the definition (and theorems derived thereon). Avoid wordings such as “tends to nothing”. A limit is a number (or $\infty$ or $-\infty$), or it doesn't exist. Functions don't “tend to”. The form of a limit can be “indeterminate”, which simply means you cannot just plug in a value to compute it.
A constant function always has a limit at every accumulation point of its domain. Indeed, if $f(x)=c$ for all $x$ in the stated domain $D$, given $\varepsilon>0$, we can take $\delta=1$ and, for $0<|x-c|<\delta$, $x\in D$, we certainly have $|f(x)-c|<\varepsilon$, because $|f(x)-c|=|c-c|=0$.
Instead of $\delta=1$, you can take your favorite positive number, it doesn't matter.
If the limit at $\infty$ or $-\infty$ makes sense for the particular domain $D$, then also these limits will be $c$.
Your case is $D=\mathbb{R}$ and $c=0$.
A: The limit of a constant $c$ is trivially equal to that constant $c$.
$$
\lim_{x\to 2}(0\times x)=\lim_{x\to 2}0=0
$$
The use (or lack thereof) of infinity is irrelevant. Suppose $x\to\infty$ instead of $x\to 2$. The answer remains unchanged.
$$
\lim_{x\to \infty}(0\times x)=\lim_{x\to \infty}0=0
$$
You are correct in the sense that $\lim_{x\to 2}(0\times x)$ is a valid limit. You are incorrect in your reasoning because infinity has nothing to do with this particular limit's validity.
