Limit as x goes to negative infinity
$\displaystyle (3x^2 + 1)^-2$
(Last part is raised to the -2 power not minus 2)
The key is to know that for $\displaystyle n>0$ is $\displaystyle \frac1{t^n}\to0$ as $\displaystyle t\to+\infty.$
Now, in your limit put $\displaystyle t=-x$ so the limit equals $\displaystyle \lim_{t\to+\infty}\frac1{(3t^2+1)^2},$ so for large $\displaystyle t$ is $\displaystyle \frac{1}{16t^{4}}<\frac{1}{\left( 3t^{2}+1 \right)^{2}}<\frac{1}{9t^{4}},$ and the squeeze theorem finishes the problem.