Limit of (Cosine (X) - 1) over X at Zero/Proof 2

Theorem

 * $\displaystyle \lim_{x \to 0} \frac {\cos \left({x}\right) - 1} {x} = 0$

Proof
This proof assumes the truth of the Derivative of Cosine Function:

We have that:


 * From Cosine of Zero is One: $\cos 0 = 1$
 * From Derivative of Cosine Function: $D_x \left({\cos x}\right) = - \sin x$ and by Derivative of Constant: $D_x \left({-1}\right) = 0$. So by Sum Rule for Derivatives $D_x \left({\cos x - 1}\right) = - \sin x$
 * By Sine of Zero is Zero, $\sin 0 = 0$
 * From Derivative of Identity Function: $D_x \left({x}\right) = 1$.

Thus L'Hôpital's Rule applies and so $\displaystyle \lim_{x \to 0} \frac {\cos x - 1} x = \lim_{x \to 0} \frac {-\sin x} 1 = \frac {-0} 1 = 0$.