How to compute $a\cdot \lim\limits_{n \to \infty}n?$ How to compute $$a\cdot \lim_{n \to \infty}n?$$
I think such an expression is not valid. Though $\lim\limits_{n \to \infty}n=\infty$, which is not a concrete real number, we can not put it into a multiply operation.
But WolframAlpha outputs confusing results. For example, it shows that$$ 1\cdot \lim_{n \to \infty}n=\infty,~~~0\cdot \lim_{n \to \infty}n=\text{undefined}.$$
How to comprehend the difference here?
 A: Because, for WolframAlpha, $1\times\infty=\infty$, whereas $0\times\infty$ is undefined.
A: A restricted multiplication (i.e. something which satisfies all the multiplication laws but cannot be defined for every pair of extended real numbers) applies with the extended real number line. In that system, indeed $ 1 \cdot \infty = \infty$ but $0 \cdot \infty$ is actually not defined.
For example, see here : https://en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations
So what Wolfram alpha is doing is assuming that $\lim_{n \to \infty} n = \infty$, and then performing the multiplication as in the extended real line. This reflected by the fact that the engine gives $1 \cdot \infty = \infty$ and $0 \cdot \infty$ as undefined.
A: In the reals ($\mathbb R$),
$$\lim_{n\to\infty}n$$
is undefined, as it cannot be expressed by a real number. Then any expression where it appears is undefined as well.
In the extended reals ($\overline{\mathbb R}=\mathbb R\cup\{-\infty,\infty\}$),
$$\lim_{n\to\infty}n=\infty.$$
Then by the computation rules in the extended reals
$$1\cdot\infty=\infty$$ and $$0\cdot\infty$$ is undefined.
