Fake proof that $1=0$ using the product law for limits

Context: I was looking at Is it true that $0.999999999\ldots=1$? Then I started thinking: if $0.999999... = 1$, then $0.000..001 = 1/1000...000 = 0$. Then if we multiply both side by $1000...000$, we will have $1 = 0$?

To rephrase it in a slightly more formal way as below. Suppose we multiply both sides of the equality $$\lim_{n\to\infty}\frac{1}{n} = 0$$ by $n$: \begin{align} n \cdot \lim_{n \to \infty}\frac{1}{n}&=n\cdot 0 \\[5pt] \lim_{n \to \infty}1 &= 0 \end{align} So $1=0$. I'm sure I am wrong, but what exactly is wrong with my reasoning?

Edit After some kind suggestion and answers from Terasa, Martin and Joe, I'd like to link a related helpful question below. The only slight difference is that question confused about the infinite sum and limits, while I confused between the product and limits!

What is wrong with the argument $1 = \lim_{n\to \infty} n/n = \lim_{n\to\infty} (1/n+1/n+\dotsb+1/n) = 0 $?

Solution 1:

In general, we have the limit law $$ k\lim_{n \to \infty}f(n)=\lim_{n \to \infty}kf(n) \, , $$ where $k$ is any constant (provided, of course, that the limit exists). However, it does not make sense to write $$ n\lim_{n \to \infty}\frac{1}{n}=\lim_{n \to \infty}1 \, . $$ The reason for this is quite subtle: the $n$ inside the limit operator is a dummy variable. The only purpose it is serving is to tell us that we are finding the limit of the function $f$ given by $f(x)=1/x$. The $n$ outside the limit operator is not a dummy variable.* So your confusion is occurring because of a notational conflict. If you write the limit as $$ n\lim_{x \to \infty}\frac{1}{x} \, , $$ where you are no longer using $n$ in two different ways, then we do indeed have $$ n\lim_{x \to \infty}\frac{1}{x} = \lim_{x \to \infty}\frac{n}{x} $$ for any real number $n$. But this does not imply that $1=0$.

*One way to check whether a variable is a 'dummy' is to consider whether you can substitute a value in for it. If you consider the limit $$ \lim_{x \to \infty}\frac{n}{x} $$ then it makes sense to plug in $n=5$, say, but plugging in $x=5$ would be absurd. That's because $1/x$ represents a function. More formally, it can be viewed as a shorthand for the function $f$ such that $$ f(x)=\frac{1}{x} $$ for all $x\neq0$.