How do you use the Squeeze Theorem to find $\lim \dfrac{\sin x}{x}$ as $x$ approaches zero?

Hint: We recall squeeze theorem $g\left( x \right)\le f\left( x \right)\le h\left( x \right),\displaystyle \lim_{x \to a}g\left( x \right)=L=\displaystyle \lim_{x \to a}h\left( x \right)$then $\displaystyle \lim_{x \to a}f\left( x \right)=L$ for any $a$ in the common domain of $f,g,h$. We draw triangle ABC with $\angle B={{90}^{\circ }}$ with $\angle CAB=x$radian. We take $AB=1$ and $PQ||BC$. We take $PQ=\sin x,BD=\tan x$. We use the inequality of areas that are the area of triangle APB $\le $ area of sector APQ $\le $ area of triangle ABC. We divide $\dfrac{1}{2}\sin x$ to get the candidates for $g\left( x \right),h\left( x \right)$.

Complete step by step solution:
We know from squeeze theorem or sandwich theorem that if there single variable real valued function $g\left( x \right),f\left( x \right),h\left( x \right)$ with following conditions $g\left( x \right)\le f\left( x \right)\le h\left( x \right)$ and $\displaystyle \lim_{x \to a}g\left( x \right)=L=\displaystyle \lim_{x \to a}h\left( x \right)$ for any $a\in I$ where $I$ is an interval in the common domain of $f,g,h$; then we have $\displaystyle \lim_{x \to a}f\left( x \right)=L$.Here $g\left( x \right),h\left( x \right)$ are respectively called upper and lower bounds of $f\left( x \right)$.\[\]
We are asked to evaluate $\lim \dfrac{\sin x}{x}$ as $x$ approaches zero in the question. So we have$f\left( x \right)=\dfrac{\sin x}{x}$. We need $g\left( x \right),h\left( x \right)$ . Let us consider triangle ABC with $\angle B={{90}^{\circ }}$ with $\angle CAB=x$radian where $x$ is a very small positive angle. We take $AB=1$unit. We drew arc AB which cuts AC at P. We draw PQ such that $PQ||BC$. We take $BC=\tan x,PQ=\sin x$. \[\]
 Now we have from above construction
\[\begin{align}
  & \text{Area of triangle APQ}\le \text{Area of sector APB}\le \text{Area of triangle ABC} \\
 & \Rightarrow \dfrac{1}{2}\times AB\times PQ\le \dfrac{1}{2}\times {{\left( AB \right)}^{2}}\times m\angle CAB\le \dfrac{1}{2}\times AB\times BC \\
 & \Rightarrow \dfrac{1}{2}\times 1\times \sin x\le \dfrac{1}{2}\times {{1}^{2}}\times x\le \dfrac{1}{2}\times 1\times \tan x \\
 & \Rightarrow \dfrac{1}{2}\sin x\le \dfrac{1}{2}x\le \dfrac{1}{2}\tan x \\
\end{align}\]
We divide the above step by $\dfrac{1}{2}\sin x$. The inequality will remain changed since $x > 0\Rightarrow \sin x > 0$. So we have
\[\begin{align}
  & \Rightarrow \dfrac{\dfrac{1}{2}\sin x}{\dfrac{1}{2}\sin x}\le \dfrac{\dfrac{1}{2}x}{\dfrac{1}{2}\sin x}\le \dfrac{\dfrac{1}{2}\tan x}{\dfrac{1}{2}\sin x} \\
 & \Rightarrow 1\le \dfrac{x}{\sin x}\le \dfrac{\tan x}{\sin x} \\
\end{align}\]
We put $\tan x=\dfrac{\sin x}{\cos x}$ and simplify the above step to have;
\[\Rightarrow 1\le \dfrac{x}{\sin x}\le \dfrac{1}{\cos x}\]
We take reciprocal of the above step to have;
\[\Rightarrow \cos x\le \dfrac{\sin x}{x}\le 1\]
Now we have the required functions $g\left( x \right)=\cos x,h\left( x \right)=1$ such that
\[\begin{align}
  & \displaystyle \lim_{x \to 0}g\left( x \right)=\displaystyle \lim_{x \to 0}\cos x=\cos 0=1 \\
 & \displaystyle \lim_{x \to 0}h\left( x \right)=\displaystyle \lim_{x \to 0}1=1 \\
\end{align}\]
So by squeeze theorem we have the required limit as
\[\displaystyle \lim_{x \to 0}f\left( x \right)=\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1\]

Note: We note that the area of triangle is $\dfrac{1}{2}\times $base$\times $height and the area of circular sector with radius $r$ and $\theta $radian is $\dfrac{1}{2}\times \theta \times {{r}^{2}}$. We must be clear of confusion that the area of the circular sector with $\theta $ in degree is $\dfrac{\theta }{360}\times {{r}^{2}}$. We also note that construction follows the definition of sine as the ratio of opposite side to hypotenuse and definition of tangent as the ratio of opposite side to adjacent side. We can use squeeze theorem at the endpoints of interval $I$ with only left or right hand limits.