Answer
0
Work Step by Step
We have to determine the limit:
$L=\lim\limits_{x \to 0}\dfrac{\cos x-1}{x}$
Multiply both numerator and denominator by $\cos x+1$:
$L=\lim\limits_{x \to 0}\dfrac{\cos x-1}{x}\cdot\dfrac{\cos x+1}{\cos x+1}=\lim\limits_{x \to 0}\dfrac{\cos^2 x-1}{x(\cos x+1)}$
Use the identity:
$\sin^2 x+\cos^2 x=1$
$L=\lim\limits_{x \to 0}\dfrac{-\sin^2 x}{x(\cos x+1)}$
$=-\lim\limits_{x \to 0}\dfrac{\sin^2 x}{x^2}\cdot\dfrac{x}{\cos x+1}=-\lim\limits_{x \to 0}\left(\dfrac{\sin x}{x}\right)^2\cdot \dfrac{x}{\cos x+1}$
Use $\lim\limits_{x \to 0} \dfrac{\sin x}{x}=1$
$L=-\lim\limits_{x \to 0} \dfrac{x}{\cos x+1}=-\dfrac{0}{1+1}=0$
