How do I solve $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3-1})$ indeterminate limit without the L'hospital rule? I've been trying to solve this limit without L'Hospital's rule as homework. So I tried rationalizing the denominator and numerator but it didn't work.
My best was: $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3-1}) = \lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{x^2+x+1-2}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{(x-1)(x+2)}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{x+2}{(x^2+x+1)}) = 3 ???$

Partial fractions save the day:
$$\frac{2}{x^3-1} = \frac{a}{x-1} + \frac{bx+c}{x^2+x+1}$$
You get $a=\frac{2}{3}$, so:
$$\frac{2}{x^3-1} =\frac{2}{3}\frac{1}{x-1} + \frac{bx+c}{x^2+x+1}$$
Now $$\lim_{x\to 1}\frac{bx+c}{x^2+x+1} = \frac{b+c}{3}.$$
So you only need to compute $$\lim_{x\to 1} \left(\frac{1}{x-1}-\frac{2}{3}\frac{1}{x-1}\right)=\lim_{x\to 1}\frac{1}{3}\frac{1}{x-1}$$
