Recall that $$|a| = \begin{cases} a, & \mbox{if } a \ge 0 \\ -a, & \mbox{if } a < 0. \end{cases} $$
Using this definition you should be able to use normal limit techniques ($\epsilon-\delta$ or what have you)
Notice, of course, that your limit does not exist as $x$ approaches zero.
Since the numerator and denominator is zero at $1$, let's factor out $(x-1)$ from both of them to get an idea how the function behaves around $1$.
The fraction equals $\dfrac{(3x^3-5x^2-5x-5)(x-1)}{(x^2-1)(x-1)}=\dfrac{3x^3-5x^2-5x-5}{x^2-1}$.
At $x=1$, the numerator equals $-12$. So for values around and very close to $1$, the numerator stays near $-12$.
The denominator however, is negative for $x<1$ and is positive for $x>1$. Thus, as $x$ approaches $1$ from the left, $x^2-1$ takes on values like $-0.1,-0.01,-0.001,\ldots$ while the numerator remains close to $-12$. Hence, the fraction is positive and becomes arbitrarily large as $x\to 1^{-}$.
Similarly, as $x\to 1^{+}$, the denominator is positive and becomes small while the numerator remains near $-12$ so that your expression here approaches $-\infty$.
Best Answer
When the denominator approaches zero and the numerator approaches a constant you have a vertical asymptote. So you have the function f: $$f(x) = \frac{x}{x^2-1}=\frac{x}{(x-1)(x+1)}$$ Where there are vertical asymptotes at $x=1$ and $x = -1$. You just need to analyse whether the function is positive or negative close to these values to see if the function is approaching positive or negative infinity.
For the specific limit as $x$ approaches 1+, the numerator will be positive ($x>0$) and the demoniator will be positive ($x-1)(x+1) > 0$ for the range $x>1$, so this limit is positive infinity.