Squeeze Theorem and the Limit of sin(x)/x

Using the formula \(\text{Area of Triangle} = \frac 12 \text{(base)}\cdot \text{(height)}\), we get that the area of the small triangle is

\begin{align} \operatorname{Area}(A) = \frac12 \cos\theta\sin\theta. \end{align}

Recall that the fraction of the area taken up by a circle wedge is \(\theta/2\pi\):

This means that

\begin{align} \operatorname{Area}(B) = \frac{\theta}{2}. \end{align}

From the diagram, it’s

\begin{align} \operatorname{Area}(C) = \frac 12 \cdot \tan\theta. \end{align}

Based on the construction of these three shapes in the original picture, we see that \(A\) sits inside \(B\) and \(B\) sits inside \(C\). Thus

\begin{align}\label{eq:areas} \operatorname{Area}(A) \leq \operatorname{Area}(B) \leq \operatorname{Area}(C). \end{align}

Substituting the formulas we found in for the areas in inequality \eqref{eq:areas}, we get

\begin{align}\label{eq:halves} \frac12 \cos\theta\sin\theta \leq \frac{\theta}{2} \leq \frac12 \tan\theta. \end{align}

Multiply by \(2\) to get

\begin{align}\label{eq:doubled} \cos\theta\sin\theta \leq \theta \leq \tan\theta = \frac{\sin\theta}{\cos\theta}. \end{align}

Now we make some restrictions. We’re eventually going to be applying the squeeze theorem at \(\theta = 0\), so we may as well restrict our possible values of \(\theta\). Let’s say that \(-\pi/2 < \theta < \pi/2\); if you look at the statement of the squeeze theorem, we have chosen \(\epsilon = \pi/2\). Now we’re working in the interval \(\left(-\frac\pi2, \frac\pi2\right)\).

Note: We actually already implicitly made this restriction to \(\theta\). If \(\theta\) was any larger, then the triangles wouldn’t have angles that sum to \(\pi\) radians.

I’d like to divide by \(\sin\theta\), but in order to control what happens to the inequalities, I need to break into the cases \(\sin\theta > 0\) and \(\sin\theta < 0\).

Case 1 (\(\theta > 0\)). This means \(\sin\theta > 0\) as well, so I can divide by it without affecting the inequalities. Dividing everything in \eqref{eq:doubled} by \(\sin\theta\) gives me

\begin{align}\label{eq:case1-divided} \cos\theta \leq \frac{\theta}{\sin\theta} \leq \frac{1}{\cos\theta}. \end{align}

Now I’ve got \(\frac{\theta}{\sin\theta}\), the opposite of what I want. Can I flip everything? Again, I need to verify that the signs behave well. Recall that, when \(a\) and \(b\) have matching signs, we get

\begin{align} a < b \iff \frac{1}{a} > \frac{1}{b} \end{align}

(play with some positive numbers to convince yourself of this if it’s unfamiliar). As long as \(0 < \theta < \pi/2\), as we have assumed is the case, then everything in these inequalities is positive. Taking the reciprocal of everything in \eqref{eq:case1-divided} gives me

\begin{align} \frac{1}{\cos\theta} \geq \frac{\sin\theta}{\theta} \geq \cos\theta, \end{align}

and magically, we have arrived at upper and lower bounds for \(\frac{\sin\theta}{\theta}\) on the interval \(\theta\in \left(0,\frac{\pi}{2}\right)\)!

Case 2 (\(\theta < 0\)). If the angle is negative, then the shapes we drew flip down into quadrant 4 of the plane. The formulas we used for area then include a negative side length, so our inequalities in \eqref{eq:halves} flip! This gives us

\begin{align} \frac12 \cos\theta\sin\theta \geq \frac{\theta}{2} \geq \frac12 \tan\theta. \end{align}

I can repeat the steps from before, except \(\theta < 0\) implies \(\sin\theta < 0\) too, so when I divide by it I must flip all the inequalities again:

\begin{align}\label{eq:case2-divided} \cos\theta \leq \frac{\theta}{\sin\theta} \leq \frac{1}{\cos\theta}. \end{align}

We’re in the interval \(\theta\in \left(-\frac{\pi}{2},0\right)\). Here, \(\cos\theta\) is positive, \(\sin\theta\) is negative and \(\theta\) is negative. This means \(\sin\theta/\theta\) is positive, so all sides of the inequalities in \eqref{eq:case2-divided} have matching sign. According to the rule from Case 1, I therefore flip the inequalities one more time when I take reciprocals, leaving me with

\begin{align} \frac{1}{\cos\theta} \geq \frac{\sin\theta}{\theta} \geq \cos\theta. \end{align}

This is the same pair of lower and upper bounds as in the other case.

We’re now ready to apply the squeeze theorem. Set \(g(\theta) = \cos\theta\) and \(h(\theta) = \frac{1}{\cos\theta}\). What we have just shown with a lovely combination of geometry and algebra is that, whenever \(\theta \in \left(-\frac\pi2,\frac\pi2\right)\), we have

\begin{align} g(\theta) \leq \frac{\sin\theta}{\theta}\leq h(\theta). \end{align}

Taking limits, we get

\begin{align} \lim_{\theta\to 0} g(\theta) = \lim_{\theta\to 0} \cos\theta = \cos(0) = 1, \end{align} \begin{align} \lim_{\theta\to 0} h(\theta) = \lim_{\theta\to 0} \frac{1}{\cos\theta} = \frac{1}{\cos(0)} = 1, \end{align}

and hence by the squeeze theorem we get

\begin{align} \lim_{\theta\to 0}\frac{\sin\theta}{\theta} = 1. \end{align}

We’re done! Here’s a last graph to illustrate that these bounds do indeed work, in case you don’t trust the algebra.