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

Theorem

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

Proof
We have that:


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