Answer
$\dfrac{7}{2}$
Work Step by Step
Write the Taylor series for the function as: $\sqrt[3] {1-6x}=1-2x-4x^2+......$ and $(1+x)^{-2}=1-2x+3x^2+......$
Now, we have:
$\lim\limits_{x \to 0} \dfrac{(1+x)^{-2}-\sqrt[3] {1-6x}}{2x^2}=\lim\limits_{x \to 0} \dfrac{1-2x+3x^2+.....-(1-2x-4x^2+......)}{2x^2}\\=\lim\limits_{x \to 0} \dfrac{3x^2+4x^2}{2x^2} \\=\dfrac{7}{2}$
