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

Proof of Corollary
We have the inequality


 * $1 \le \dfrac \theta {\sin \theta} \le \dfrac 1 {\cos \theta}$

for all $\theta \in \left({-\dfrac \pi 2 .. 0}\right) \cup \left({0 .. \dfrac \pi 2}\right)$.

Taking the limit of the leftmost term and the rightmost term:


 * $\displaystyle \lim_{\theta \to 0} \ 1 = 1$


 * $\displaystyle \lim_{\theta \to 0} \frac{1}{\cos\theta} = 1$

So by the Squeeze Theorem:


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