Solve $\lim\limits_{x \to 1} \frac{2x-2}{(x^3-x)^2}$ without L'Hôpital's rule Find limit without L'Hôpital's rule: $\lim\limits_{x \to 1} \frac{2x-2}{(x^3-x)^2}$
I'm learning limits and I'm stuck with this example. So far, I tried modify denominator but I have no luck with it.

$$\lim\limits_{x \to 1} \frac{2x-2}{(x^3-x)^2}=\lim\limits_{x \to 1} \frac{2(x-1)}{x^2(x^2-1)^2}=\lim\limits_{x \to 1} \frac{2(x-1)}{x^2(x-1)^2(x+1)^2}$$
Therefore $\lim\limits_{x \to 1^+} \frac{2}{x^2(x-1)(x+1)^2}=+\infty$
and $\lim\limits_{x \to 1^-} \frac{2}{x^2(x-1)(x+1)^2}=-\infty$
