TITLE: What is the limit of zero times x, as x approaches infinity?
QUESTION [9 upvotes]: I am having difficulty determining is the solution for the following problem:
$$\displaystyle \lim_{x \rightarrow \infty}\left( x \times 0 \right)$$
To clarify, this question assumes ${0}$ is a constant and is absolutely zero ("true zero"), and not another figure approaching or is approximately zero ("near zero"). Thus, the question is not asking what "near zero" times "near infinity" is.
I know that ${\infty *0}$ is undefined, however my difficulty is that I'm unsure whether the answer to the problem is undefined because ${\infty *0}$ is undefined.
From my understanding, a limit does not ever 'reach' infinity - it only approaches infinity, thus there are a rational amount of numbers. As  ${x\cdot 0=0}$, when x is not ${\infty}$, it seems to me that in all cases of $x$ approaching infinity the answer could also be ${0}$.

REPLY [6 votes]: When you evaluate a limit you don't actually have to worry about the value of the function at the index. In other words, when you evaluate $$\lim_{x\rightarrow x_0}f(x),$$
it is not necessary for $f(x_0)$ to be defined. A limit tells you the 'best guess' for $f(x_0)$ based on its surroundings. Take the function $$f(x)=\frac{x-3}{x-3};$$
it's not hard to see that $f$ evaluates to 1 for any $x$ except for $x=3$, where it is undefined. Based on the surroundings of $x=3$ it is possible though to give a sensible value to $f(3)$. You can find this value using a limit:
$$\lim_{x\rightarrow 3}\frac{x-3}{x-3}=1$$
The limit yields 1 since you can get $f(x)$ arbitrarily close to 1 by picking $x$ arbitrarily close to $3$ (in this case it doesn't actually matter if you get closer but still).
To tie this back to your question: the limit is zero, because for every $x$ smaller than infinity you have that $x\cdot 0$ is arbitrarily close to $0$. It doesn't matter whether $\infty\cdot 0$ makes sense or not. That is why limits are so useful.