Proof: limit sinθ/θ=1 – Sam Artigliere's Blog

Statement:

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

Proof:

The circle in figure 1 is a unit circle. $OA$ and $OB$ are radii. Therefore,

$OA=OB=1$
$AD=\sin\theta$
$\tan\theta=\frac{BC}{OB}\,\,\text{but}\,\,OB=1\,\text{so}\,\tan\theta=BC$

We need to calculate the area of the arc between $OB$ and $OA$ swept out over $\angle\theta$ . this calculation goes as follows:

The circumference of the entire circle is given by $2\pi r$ .
The length of the arc between $A$ and $B$ is $\theta \cdot r$ (where $\theta$ is expressed in radians)
The fraction of the circle made up from the arc between $A$ and $B$ , therefore, is given by $\frac{\theta r}{2\pi r}=\frac{\theta}{2\pi}$
The area of the entire circle is equal to $\pi r^2$
The fraction of the area of the circle made from the arc between $A$ and $B$ , then, equals $\frac{\theta}{2\pi} \cdot \pi r^2=\frac{\theta}2 \cdot r^2=\frac{\theta}2 \cdot 1^2=\frac{\theta}2$ so,
$\text{Area}_{arc AOB}=\frac{\theta}2$

Next, we need to calculate the areas of $\triangle AOB$ and $\triangle COB$ .

$\text{Area}_{\triangle AOB}$ :

The area of a triangle is $\frac12 \text{base}\cdot\text{height}$ .
The base is the radius of the circle, 1.
The height is $\sin\theta$ .
Therefore, $\text{Area}_{\triangle AOB}=\frac{\sin \theta}{2}$

$\text{Area}_{\triangle COB}$ :

The area of a triangle is $\frac12 \text{base}\cdot\text{height}$ .
The base is the radius of the circle, 1.
The height is $\tan\theta$ . Why? Because $\tan=\frac{\text{opposite}}{\text{adjacent}} = \frac{\text{Height}_{COB}}{\text{radius of circle}}=\frac{\text{Height}_{COB}}{1}=\text{Height}_{COB}$ .
Therefore, $\text{Area}_{\triangle COB}=\frac{\tan \theta}{2}$

Just by looking at the diagram, we can see that:

$\text{Area}_{\triangle AOB}\leq\text{Area}_{arc AOB}\leq\text{Area}_{\triangle COB}$

Thus,

$\frac{\tan \theta}{2}\geq\frac{\theta}2\geq\frac{\sin \theta}{2}=\frac{\sin \theta}{2\cos\theta}\geq\frac{\theta}2\geq\frac{\sin \theta}{2}$

Multiply through by 2:

$\frac{\sin \theta}{\cos\theta}\geq\theta\geq\sin \theta$

Divide through by $\sin\theta$ :

$\frac{\cancel{\sin \theta}}{\cancel{\sin\theta}\cos\theta}\geq\frac{\theta}{\sin\theta}\geq\frac{\cancel{\sin \theta}}{\cancel{\sin\theta}}=\frac{1}{\cos\theta}\geq\frac{\theta}{\sin\theta}\geq 1$

Take the reciprocal of this equation:

$\cos\theta\leq\frac{\sin\theta}{\theta}\leq 1$

Notice that we changed the $\geq$ sign to $\leq$ . Why? Here’s an example:

$\frac12\geq\frac13\,\,\rightarrow\,\, 2\leq3$

The squeeze theorem states that:

If $f(x) \leq g(x) \leq h(x)$ and
If $\displaystyle\lim_{x \to c} f(x) = L$ and $\displaystyle\lim_{x \to c} h(x) = L$
Then $\displaystyle\lim_{x \to c} g(x) = L$

To apply the squeeze theorem to our problem, let

$x=\theta$
$c=0$
$L=1$

And take the limit as $\theta\to 0$ of the equation $\cos\theta\leq\frac{\sin\theta}{\theta}\leq 1$ . We get:

$\begin{array}{ccccc} \displaystyle\lim_{\theta \to 0} \cos\theta &\leq& \displaystyle\lim_{\theta \to 0} \frac{\sin\theta}{\theta} &\leq& \displaystyle\lim_{\theta \to 0} 1 \\ \, &\,& \,&\,&\, \\ 1 &\leq& \displaystyle\lim_{\theta \to 0} \frac{\sin\theta}{\theta} &\leq& 1 \end{array}$

Therefore,

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