Limit of Sine of X over X/Corollary

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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:

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

$\blacksquare$