Answer
0
Work Step by Step
We have to determine:
$L=\lim\limits_{x \to 0}\dfrac{\cos x-1}{x}$
Use the half-angle formula:
$\sin^2 x=\dfrac{1-\cos 2x}{2}$
$L=-\lim\limits_{x \to 0}\dfrac{2\sin^2\dfrac{x}{2}}{x}=-2\lim\limits_{x \to 0}\dfrac{\sin\dfrac{x}{2}}{\dfrac{x}{2}}\cdot \dfrac{\sin\dfrac{x}{2}}{2}$
Use the fact that $\lim\limits_{x \to 0}\dfrac{\sin x}{x}=1$
$L=-2\lim\limits_{x \to 0}\dfrac{\sin\dfrac{x}{2}}{2}=-2\cdot \dfrac{0}{2}=0$
