Limits of $f(x)=x-x$ It's obvious that $f(x)=x-x=0$. But what exactly happens here?
You have a function $f(x)=x-x$ and you have to calculate the limits when $x\to \infty$
This'll be like this:
$$\lim\limits_{x\to \infty}f(x)=\infty - \infty$$ That's an indetermination, and you have to multiply both sides with the conjugate of $f(x)$, which is equal to $x+x$.
\begin{align}
f(x)&=x-x\\
&=\frac{(x-x)(x+x)}{x+x}\\
&=\frac{x^2-x^2}{x+x}
\end{align}
If we do the limits now the answer is going to be: $$\lim\limits_{x\to \infty}f(x)=\frac{\infty - \infty}{\infty+\infty}$$ Which it's another type of indetermination(I think).

What happens here?Can there be an error multiplying with it's conjugate in both sides? Is there another case like this? Or am I completely wrong?


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 $x-x=0$.
