The Limit of sinx/x
Formula
$$ \lim \limits_{x \to 0} \dfrac{\sin x}{x} = 1 $$
Proof
Suppose we are given a sector $OAB$ of radius $1$. Let $H$ be the foot of the perpendicular dropped from point $B$ to the segment $\overline{OA}$. And let $C$ be the intersection point obtained by extending the segments $\overline{OB}$ and $\overline{OA}$.
Then the lengths of each segment are as follows.
$$ \overline{OH} = \cos \theta, \qquad \overline{BH} = \sin \theta, \qquad \overline{AC} = \tan θ $$
The area of triangle $OBH$ is $\dfrac{1}{2} \times \overline{OH} \times \overline{BH} = \dfrac{1}{2} \cos\theta \sin\theta$. The area of sector $OAB$ is $\dfrac{1}{2} \theta r^{1} = \dfrac{1}{2} \theta$. Also, the area of triangle $OAC$ is $\dfrac{1}{2} \overline{OA} \times \overline{AC} = \dfrac{1}{2} \tan\theta$. Among the areas of these three figures, the following inequality holds.
$$ \dfrac{1}{2} \cos\theta \sin\theta \lt \dfrac{1}{2} \theta \lt \dfrac{1}{2} \tan\theta \implies \cos\theta \sin\theta \lt \theta \lt \dfrac{\sin\theta}{\cos\theta} $$
Taking the reciprocal of the above inequality, we obtain the following.
$$ \dfrac{\cos\theta}{\sin\theta} \lt \dfrac{1}{\theta} \lt \dfrac{1}{\sin\theta \cos\theta} $$
Since $\theta$ is an acute angle, multiplying each term by $\sin \theta$ does not change the direction of the inequality.
$$ \cos\theta \lt \dfrac{\sin\theta}{\theta} \lt \dfrac{1}{\cos\theta} $$
Taking the limit as $\theta \to 0$ here, by the squeeze theorem we obtain the following.
$$ \begin{align*} && \lim\limits_{\theta \to 0} \cos\theta \lt \lim\limits_{\theta \to 0} \dfrac{\sin\theta}{\theta} \lt \lim\limits_{\theta \to 0} \dfrac{1}{\cos\theta} \\ \implies && 1 \lt \lim\limits_{\theta \to 0} \dfrac{\sin\theta}{\theta} \lt 1 \\ \implies && \lim\limits_{\theta \to 0} \dfrac{\sin\theta}{\theta} = 1 \end{align*} $$
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