<s>[INST] finding values of $a$ for which the limit exists I've been banging my head all day long over this question, would apprecate some help :)

$$f(x) = \frac{1}{x^2-1} - \frac{a}{x^3-1},\qquad a\in\mathbb{R}.$$
  For which values of $a$ does $\lim\limits_{x \to 1} f( x)$ exist?

Also it was requested to use only the $\epsilon$-$\delta$ definition.
 [/INST] Both denominators are $0$ when $x=1$; therefore $(x-1)$ is a factor of both.  So factor them, then find the common denominator:
$$
\begin{align}
& \frac{1}{x^2-1} - \frac{a}{x^3-1} = \frac{1}{(x-1)(x+1)} - \frac{a}{(x-1)(x^2+x+1)} \\  \\
& = \frac{x^2+x+1}{(x-1)(x+1)(x^2+x+1)} - \frac{a(x+1)}{(x-1)(x+1)(x^2+x+1)}
\end{align}
$$
The limit can exist only if the numerator
$$
x^2+x+1 - a(x+1)
$$
is $0$ when $x=1$.  And it is equal to $3-2a$ when $x=1$.
Can you take it from there?
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