Limit of Sine of X over X at Zero/Corollary

Theorem

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

Proof
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:


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


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

So by the Squeeze Theorem:


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