Answer
$\lim\limits_{y \to 0}\frac{(5y^{3}+8y^{2})}{(3y^{4}-16y^{2})}=-0.5=-\frac{1}{2}$
Work Step by Step
Simplify:
$\frac{(5y^{3}+8y^{2})}{(3y^{4}-16y^{2})} = \frac{y^{2}(5y+8)}{y^{2}(3y^{2}-16)} = \frac{(5y+8)}{(3y^{2}-16)}$
Now,
$\lim\limits_{y \to 0}\frac{(5y^{3}+8y^{2})}{(3y^{4}-16y^{2})} = \lim\limits_{y \to 0}\frac{(5y+8)}{(3y^{2}-16)} = -\frac{8}{16} = -\frac{1}{2} = -0.5$
