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