Limit of Sine of X over X/Proof 2

Theorem

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

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:

$\displaystyle \lim_{x \mathop \to 0} \frac {\sin x} x = \lim_{x \mathop \to 0} \frac {D_x \left({\sin x}\right)} {D_x \left({x}\right)} = \lim_{x \mathop \to 0} \frac {\cos x} 1 = \frac 1 1 = 1$

$\blacksquare$