# Limit of (Cosine (X) - 1) over X/Proof 3

## Theorem

$\displaystyle \lim_{x \mathop \to 0} \frac {\map \cos x - 1} x = 0$

## Proof

 $\displaystyle \lim_{x \mathop \to 0} \ \frac {\cos x - 1} x$ $=$ $\displaystyle \lim_{x \mathop \to 0} \ \frac {\left({\cos x - 1}\right) \left({\cos x + 1}\right)} {x \left({\cos x + 1}\right)}$ $\displaystyle$ $=$ $\displaystyle \lim_{x \mathop \to 0} \ \frac {\cos^2 x - 1} {x \left({\cos x + 1}\right)}$ $\displaystyle$ $=$ $\displaystyle \lim_{x \mathop \to 0} \ \frac {-\sin^2 x} {x \left({\cos x + 1}\right)}$ Sum of Squares of Sine and Cosine $\displaystyle$ $=$ $\displaystyle \left({\lim_{x \mathop \to 0} \ \frac {\sin x} x}\right) \left({\lim_{x \mathop \to 0} \ \frac {-\sin x} {\cos x + 1} }\right)$ Product Rule for Limits of Functions $\displaystyle$ $=$ $\displaystyle 1 \times \left({\lim_{x \mathop \to 0} \ \frac{-\sin x} {\cos x + 1} }\right)$ Limit of Sine of X over X $\displaystyle$ $=$ $\displaystyle \frac {\lim_{x \mathop \to 0} \left({- \sin x}\right)} {\lim_{x \mathop \to 0} \left({\cos x + 1}\right)}$ Quotient Rule for Limits of Functions $\displaystyle$ $=$ $\displaystyle \frac 0 2$ $\displaystyle$ $=$ $\displaystyle 0$

$\blacksquare$