As several others have pointed out, $\infty$ is not a number. So you need to deal it with a bit of care.
To clarify your doubt, the way you have written is to look at $n \times 0$ and then let $n \rightarrow \infty$. So it is true that $$\displaystyle \lim_{n \rightarrow \infty}\left( n \times 0 \right)= 0$$
However, when people write $\infty \times 0$ usually it is a shorthand to denote the indeterminate form when some quantity tends to infinity and some other quantity tends to zero in a limiting sense i.e. expressions of the form $$\lim_{x \rightarrow 0} \left( f(x) \times g(x) \right)$$ where $\displaystyle \lim_{x \rightarrow 0} f(x) = \infty$ and $\displaystyle \lim_{x \rightarrow 0} g(x) = 0$.
(Note that $\infty$ is not a number in the conventional sense. It is just a shorthand to denote that something grows unbounded i.e. given any number your function can take a value larger than that number.)
For instance, let $f(x) = \frac{1}{x}$ as $g(x) = x$, then $f(x) \times g(x) = 1$, $\forall x \neq 0$ and hence $$\displaystyle \lim_{x \rightarrow 0} \left( f(x) \times g(x)\right) = \lim_{x \rightarrow 0} 1 = 1$$ However, $\displaystyle \lim_{x \rightarrow 0} f(x) = \infty$ and $\displaystyle \lim_{x \rightarrow 0} g(x) = 0$ and hence in this case, the indeterminate form evaluates to $1$.
The case which resembles what you have written down is when $f(x) = \frac{1}{x}$ and $g(x) = 0$. This again is an indeterminate form since $\displaystyle \lim_{x \rightarrow 0} f(x) = \infty$ and $\displaystyle \lim_{x \rightarrow 0} g(x) = 0$. However in this case, $f(x) \times g(x) = 0$, $\forall x \neq 0$ and hence $$\displaystyle \lim_{x \rightarrow 0} \left( f(x) \times g(x) \right) = 0$$
Yet another example is to look at $f(x) = \frac{1}{x}$ and $g(x) = \sqrt{x}$. This again is an indeterminate form since $\displaystyle \lim_{x \rightarrow 0} f(x) = \infty$ and $\displaystyle \lim_{x \rightarrow 0^+} g(x) = 0$. Note that $f(x) \times g(x) = \frac{1}{\sqrt{x}}$, $\forall x \neq 0$ and hence $$\displaystyle \lim_{x \rightarrow 0^+} (f(x) \times g(x)) = \displaystyle \lim_{x \rightarrow 0^+} \frac{1}{\sqrt{x}} = \infty$$
Hence, you cannot associate a unique value to $\infty \times 0$. It depends on the problem at hand. You can read more about indeterminate form here. (As always with wikipedia, read it just to get a general overall idea.)