Limit of Sine of X over X at Zero/Corollary

Theorem

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

Proof
We have the inequality:
 * $1 \le \dfrac \theta {\sin \theta} \le \dfrac 1 {\cos \theta}$

for all $\theta \in \openint {-\dfrac \pi 2} 0 \cup \openint 0 {\dfrac \pi 2}$.

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


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


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

So by the Squeeze Theorem:


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