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?
Best Answer
$\lim\limits_{x\to\infty}(x-x)$ 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 $\lim\limits_{x\to\infty}0=0$.