$Lim_{x→0}$ $\frac{\sqrt[]{cosx} -\sqrt[3]{cosx} }{sin^2x}$
= $Lim_{x→0}$ $\frac{\sqrt[]{cosx}-1 }{1-cos^2x}$- $Lim_{x→0}$ $\frac{ \sqrt[3]{cosx}-1 }{1-cos^2x}$
= $Lim_{x→0}$ $\frac{cosx-1 }{(1-cosx)(1+cosx)(\sqrt[]{cosx}+1)}$- $Lim_{x→0}$ $\frac{ cosx-1 }{(1-cosx)(1+cosx)(\sqrt[3]{cos^2x} +\sqrt[3]{cosx}+1) }$
= $Lim_{x→0}$ $\frac{-1 }{(1+cosx)(\sqrt[]{cosx}+1)}$- $Lim_{x→0}$ $\frac{ -1 }{(1+cosx)(\sqrt[3]{cos^2x} +\sqrt[3]{cosx}+1) }$ =$\frac{-1}{4}$ +$\frac{1}{6}$ =$\frac{-1}{12}$
