Answer
$\dfrac{3}{11}$
Work Step by Step
Consider: $\lim\limits_{x \to 1}f(x)=\lim\limits_{x \to 1}\dfrac{x^3-1}{4x^3-x-3}$
Need to check that the limit has an indeterminate form.
Thus, $f(1)=\dfrac{1^3-1}{4(1)^3-1-3}=\dfrac{0}{0}$
Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$
This implies:
$\lim\limits_{x \to 1}\dfrac{3x^2}{12x^2-1}=\lim\limits_{x \to 1}\dfrac{3(1)^2}{12(1)^2-1}=\dfrac{3}{11}$
