<s>[INST] 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: $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3-1}) = \lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{x^2+x+1-2}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{(x-1)(x+2)}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{x+2}{(x^2+x+1)}) = 3 ???$
 [/INST] 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.
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