Limit of Sine of X over X at Zero

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

Proof
This proof works directly from the definition of the sine function:

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Alternative Proof
This proof assumes the truth of the Derivative of Sine Function:

We have that:


 * From Basic Properties of Sine Function: $$\sin 0 = 0$$;
 * From Derivative of Sine Function: $$D_x \left({\sin x}\right) = \cos x$$. Then by Basic Properties of Cosine Function, $$\cos 0 = 1$$;
 * From Derivative of Identity Function: $$D_x \left({x}\right) = 1$$.

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