Proof: limit sinθ/θ=1

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