Is the limit of $f(n) = n-n$ zero as $n\rightarrow \infty$? I have been working on a proof which involves sums and products going to infinity. I am wondering whether the following proof of a limit is valid, and whether that result would allow me to come to another conclusion.
What is:
$$\lim \limits_{n \to \infty} f(n)\text {, where }f(n) = n-n$$
I have worked this out to be
$$\lim \limits_{n \to \infty} n-n = \lim \limits_{n \to \infty} n(1-1) = \lim \limits_{n \to \infty} n\cdot 0 = 0$$
I'm not sure whether this is the correct way of proving this limit, or whether the answer is correct. My math teacher had said that the whole limit raised a red flag in his mind, and he wasn't sure why.
If my limit is correct, though, I would like to know whether the following is also valid:
$$\lim \limits_{n \to \infty} f(n)\cdot n = 0$$

If $f(n) = n-n$, then $f(n) = 0$ for all $n$. The limit you gave is true, i.e.
$$\lim\limits_{n\to \infty} f(n) = 0$$
is correct.
Furthermore, we have that $n\cdot f(n) = n\cdot 0 = 0$ for all $n$, so the limit
$$\lim\limits_{n\to\infty}n\cdot f(n) = 0$$
is also correct.

The red flag probably stems from the well known "indeterminate form" $\infty -\infty$. This is short hand notation for the fact that knowing
$$\lim\limits_{n\to \infty} g(n) = \infty\quad\text{ and }\quad \lim\limits_{n\to\infty} h(n) = \infty $$
is not enough by itself to determine
$$\lim\limits_{n\to\infty}(g(n)-h(n)).$$
Similarly, $\infty\cdot 0$ is also an indeterminate form, i.e. knowing
$$\lim\limits_{n\to \infty} g(n) = \infty\quad\text{ and }\quad \lim\limits_{n\to\infty} h(n) = 0$$
is not enough to determine
$$\lim\limits_{n\to\infty} g(n)\cdot h(n).$$
