Limits of $f(x)=x-x$

$\lim\limits_{x\to\infty}(x-x)$ isn't indeterminate: since $x-x=0$ for all $x$, it's simply $\lim\limits_{x\to\infty}0=0$. You certainly don't have to multiply by $\frac{x+x}{x+x}$, since there's a very simple, direct way to evaluate the limit. If you do perform this unnecessarly multiplication to get $$\lim_{x\to\infty}\frac{x^2-x^2}{x+x}\;,$$ you get a second chance to realize that you can simplify the expression: $\frac{x^2-x^2}{x+x}$ is identically $0$ for $x\ne 0$, and the singularity at $x=0$ is irrelevant to the limit as $x\to\infty$, so once again you have simply $\lim\limits_{x\to\infty}0=0$.

Never treat the symbol $\infty$ as a number subject to the usual rules of arithmetic. In most limit questions that you are asked, that procedure will give you the wrong answer. It can be useful, in a question about $\lim_{x\to \infty}f(x)$, to imagine that $x$ is a very large specific number, like $10^{40}$. It is clear that if $x=10^{40}$, then $x-x=0$.

The problem with indeterminate forms is that you measure things which approach the same limit, but at possibly different rates and what you are interested is the limit of the difference.

This is why $\lim 2n-n=\infty-\infty$ is infinite, because we evaluate the limit of the difference, not the difference of limits.

On the other hand, $f(x)=x-x$ is constantly zero, the difference is constant and therefore the limit is constant too, similarly $g(x)=\frac{x}{2x}$ has a constant ratio of $\frac12$ when approaching infinity, and the same logic applies.