Answer
Converges to $-\dfrac{3}{8}$.
Work Step by Step
Let us consider that $I_n=\int_{n}^{-2} \dfrac{3}{x^3} \ dx$
After integrating, we have: $I_n=[\dfrac{-3}{2x^2}]_n^{-2} =\dfrac{3}{2n^2}-\dfrac{3}{8}$
Now, $I=\lim\limits_{n \to -\infty} I_n=\int_{-\infty}^{-2} \dfrac{3}{x^3} \ dx\\=\lim\limits_{n \to -\infty} (\dfrac{3}{2n^2}-\dfrac{3}{8})$
Since, $\dfrac{1}{n^2}\to 0$ as $n \to \infty$
$I=-\dfrac{3}{8}$
This means that the limit exist and the given integral converges to $-\dfrac{3}{8}$.
