Answer
$1$
Work Step by Step
Here, we have $\lim\limits_{x \to 0} f(0)=\lim\limits_{x \to 0}\dfrac{1-\cos x-\cos x \sin x}{\sin 0}=\dfrac{0}{0}$
This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{x \to 0} \dfrac{\sin x+\cos 2x}{\cos x}=\dfrac{\sin 0+\cos 0}{\cos 0}=1$
or, $\dfrac{0+1}{1}=1$
