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

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Theorem

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


Proof

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


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:

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

$\blacksquare$