The expression is
[tex]\dfrac{-2n(3n^2-2)^2}{(2-4n^3)^2} = \dfrac{-18n^5+24n^3-8n}{16n^6-16n^3+4}[/tex]
you didn't specify which limit you're interested in. However, these are the "interesting" limits, so you're likely looking for one of these:
- The limits at [tex]n\to\pm\infty[/tex] are zero
- The limit at [tex]n\to\frac{1}{\sqrt[3]{2}}^-[/tex] is [tex]-\infty[/tex]
- The same limit, from the right, is zero.
