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

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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:

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

$\blacksquare$