Answer
$$-2$$
Work Step by Step
The Taylor series can be written as follows: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-...........$ and $\cos x=1-\dfrac{x^2}{2}+\dfrac{ x^3}{3}-....$
$$\lim\limits_{z \to 0} \dfrac{1-\cos^2 z}{\ln (1-z) + \space \sin z}=\lim\limits_{z \to 0} \dfrac{1- (1-z^2+\dfrac{z^4}{3}-.......)}{(-z-\dfrac{z^2}{2}-....)+(z-(z^3/3!)+.....} \\=\lim\limits_{z \to 0} \dfrac{1-(z^2/3)+...}{-\dfrac{1}{2}-\dfrac{z}{3}-....} \\=-2$$
