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: $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3-1}) = \lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{x^2+x+1-2}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{(x-1)(x+2)}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{x+2}{(x^2+x+1)}) = 3 ???$
On
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 $$\lim_{x\to 1} \left(\frac{1}{x-1}-\frac{2}{3}\frac{1}{x-1}\right)=\lim_{x\to 1}\frac{1}{3}\frac{1}{x-1}$$
On
this limit doesn't exist since $$\frac{1}{x-1}-\frac{2}{x^3-1}=\frac{x^2+x-1}{(x-1)(x^2+x+1)}$$
On
First note that $x^3 - 1 = (x-1)(x^2 + x + 1)$. Therefore multiply $\frac{1}{x-1}$ by 1 in the form $\frac{x^2 + x + 1}{x^2 + x + 1}$ and combine the fractions as $\frac{(x^2 + x + 1) - 2}{(x-1)(x^2 + x + 1)} = \frac{x^2 + x -1}{(x-1)(x^2 + x + 1)}$. The numerator is now going to $1 + 1 - 1 = 1$, while the denominator is going to $(1-1)(1 + 1 + 1) = (0)(3) = 0$. Notice that $1/0$ is not an indeterminant form. Notice that if we look at the limit from the right, $\lim_{x \to 1^+} \frac{x^2 + x -1}{(x-1)(x^2 + x + 1)}$, the top is going to $1$, while the bottom is going to $0$ through positive values, so the limit evaluates to $+\infty$. Checking the limit from the right in the same way, the top is still going to $1$, but the bottom is going to $0$ through negative values, so the limit evaluates to $-\infty$. The limit does not exist.
Picking up from your second step...
$$\lim_{x\to1}\frac{x^2+x+1-2}{(x-1)(x^2+x+1)}$$ $$=\lim_{x\to1}\frac{x^2+x-1}{(x^3-1)}$$
Let $x\to1$ and find the limit is unbounded or does not exist.