# Why $\lim\limits_{x\to 1}{(\frac{1}{x^2-x} - \frac{3}{x^3-1})} = (\infty - \infty)$

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?

• The first point is incorrect, since $0.9999^2 < 0.9999$, so the denominator will be negative there also. That is, you either have $-\infty - (- \infty)$ if you approach from the left, of $\infty - (+\infty)$ if you approach from the right. Therefore, this limit, on both sides, if the form $\infty - \infty$. Apr 25 '18 at 5:05
• firstly, the first term goes to $-\infty$ since $x^2 < x$. Second, what happens when you approach $1$ from the right hand side, i.e., $x = 1,0001$? Apr 25 '18 at 5:05
• You are right. I totally forgot that $0.999^2 < 0.999$ Apr 25 '18 at 5:12

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.