Concepts about limit: $\lim_{x\to \infty}(x-x)$ and $\lim_{x\to \infty}x-\lim_{x\to \infty}x$.

I know that $\infty-\infty$ is indeterminate, the how about $\lim_{x\to \infty}(x-x)$ and $\lim_{x\to \infty}x-\lim_{x\to \infty}x$?

Thank you.

No, $x-x$ is not indeterminate: it’s identically $0$, and therefore $\lim\limits_{x\to\infty}(x-x)=\lim\limits_{x\to\infty}0=0$ as well. However, $\lim\limits_{x\to\infty}x-\lim\limits_{x\to\infty}x$ is undefined: we don’t (in this setting) try to define $\infty-\infty$.

Added: Aha! I think that I just realized what the source of the confusion is. If $\lim\limits_{x\to\infty}f(x)=\infty$ and $\lim\limits_{x\to\infty}g(x)=\infty$, then the limit $$\lim_{x\to\infty}\big(f(x)-g(x)\big)\tag{1}$$ is indeterminate without information about the specific functions $f$ and $g$. Depending on $f$ and $g$, $(1)$ can be anywhere between $-\infty$ and $\infty$, or it can fail to exist altogether. We therefore speak of an indeterminate form $\infty-\infty$. Remember, though, that once you have specific functions, the limit is not indeterminate, though it may not exist. It’s just that the information that $\lim\limits_{x\to\infty}f(x)=\infty$ and $\lim\limits_{x\to\infty}g(x)=\infty$ isn’t enough to determine what it is.

This is in contrast to what we might call a ‘$2-3$’ form, a $\lim_{x\to\infty}\big(f(x)-g(x)\big)$ when it’s known that $\lim\limits_{x\to\infty}f(x)=2$ and $\lim\limits_{x\to\infty}g(x)=3$: in this case we can say with complete confidence that $$\lim_{x\to\infty}\big(f(x)-g(x)\big)=2-3=-1\;.$$

To start with, $$a=\displaystyle \lim_{x \to \infty} \left(f(x) - g(x) \right)$$ and $$b=\displaystyle \lim_{x \to \infty} f(x) - \lim_{x \to \infty} g(x)$$ denote two different things.

When $-\infty < \displaystyle \lim_{x \to \infty} f(x) < \infty$ and $-\infty < \displaystyle \lim_{x \to \infty} g(x) < \infty$, then both happen to be equal to each other.

A possible source of confusion is probably because people often use the same variable $x$ in $\displaystyle \lim_{x \to \infty} f(x)$ and $\displaystyle \lim_{x \to \infty} g(x)$ (though you need to understand that the variable tending to $\infty$ is just a dummy variable). It might be better to write the second expression as $$b=\displaystyle \lim_{x \to \infty} f(x) - \lim_{y \to \infty} g(y)$$

In your case, where $f(x) = g(x) = x$, then $$\displaystyle \lim_{x \to \infty} \left(f(x) - g(x) \right) = \displaystyle \lim_{x \to \infty} \left(x-x \right) = \lim_{x \to \infty} 0 = 0$$

However, $$\displaystyle \lim_{x \to \infty} f(x) - \lim_{y \to \infty} g(y) = \displaystyle \lim_{x \to \infty} x - \lim_{y \to \infty} y$$ doesn't exist since both $\displaystyle \lim_{x \to \infty} f(x)$ and $\displaystyle \lim_{y \to \infty} g(y)$ are $\infty$.

Here is one reason why "$\infty-\infty$" is indeterminate in this context:

Pick any $a\in\mathbb R$. Define $f(x)=x$ and $g(x)=x-a$. Then both $f(x)\to\infty$ and $g(x)\to\infty$ as $x\to\infty$, but $f(x)-g(x)\to a$. So, we should have $\infty-\infty=a$ for every $a\in\mathbb R$.

You can also define $f(x)=2x$ and $g(x)=x$ to get "$\infty-\infty=\infty$" or $f(x)=x$ and $g(x)=2x$ to get "$\infty-\infty=-\infty$".

Also, you can define $f(x)=x$ and $g(x)=x+\sin x$. Again $f(x)\to\infty$ and $g(x)\to\infty$ but $\lim_{x\to\infty}(f(x)-g(x))$ doesn't exist at all.