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