I am going through limit examples and see the following:
$$\lim\limits_{x\to 1}{(\frac{1}{x^2-x} - \frac{3}{x^3-1})} = (\infty - \infty) $$
Why is it so? As our x -> 1, we are going to have (for example) in the left side:
$$\frac{1}{0.9999^2-0.9999}$$ And here I agree that the result will -> $\infty$.
However, in the right side:
$$\frac{3}{0.9999^3-1}$$
Which will give us $-\infty$ (as denominator will be negative).
So we are going to have $\infty - (-\infty)$. Which is $\infty + \infty$ instead of $(\infty - \infty)$ as in the example solution.
Am I right and the example solution is wrong?
Note that for $0<a<1$, then $a^2<a\rightarrow a^2-a<0$, as like the case $$\frac{1}{0.9999^2-0.9999}$$will be negative infinity.
Therefore, the result would be $-\infty-(-\infty)$ which can be determined.