Limit of Sine of X over X at Zero

Theorem

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

Corollary

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

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

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