Limit of Sine of X over X at Zero/Proof 2

Proof
From Sine of Zero is Zero:
 * $\sin 0 = 0$

From Derivative of Sine Function:
 * $D_x \left({\sin x}\right) = \cos x$

Then by Cosine of Zero is One:
 * $\cos 0 = 1$

From Derivative of Identity Function:
 * $D_x \left({x}\right) = 1$

Thus L'Hôpital's Rule applies and so:
 * $\displaystyle \lim_{x \mathop \to 0} \frac {\sin x} x = \lim_{x \mathop \to 0} \frac {D_x \left({\sin x}\right)} {D_x \left({x}\right)} = \lim_{x \mathop \to 0} \frac {\cos x} 1 = \frac 1 1 = 1$