Limiting the Sine Function

We are going to start by looking at \[\lim_{x \to 0} \frac{\sin{x}}{x}\] This is one of the most important limit results and is surprisingly easy to prove. All we will need is the "squeeze method" and a pretty picture.

The Image

Here it is then. The first quadrant of the unit circle, with an extended diameter. The triangles \(\triangle{OAB}\) and \(\triangle{OCD}\) are both right-angled. With a bit of thought, you should be able to convince yourself that \(\overline{CD} = \sin{x}\) and \(\overline{AB} = \tan{x}\). These follow simply from the fact that \(\overline{OA} = \overline{OC} = 1\) since both are radii of the unit circle. The proof is based on comparing the the areas of the triangles \(\triangle{OAC}\) and \(\triangle{OCD}\), and the sector \(OAC\) of the unit circle. Easy to see that in terms of area \begin{align*} \triangle{OAC} &\le OAC \le \triangle{OAB} \\ \tfrac12 \sin{x} &\le \tfrac12 x \le \tfrac12 \tan{x} \\ \sin{x} &\le x \le \tan{x} & \text{and dividing by $\sin{x}$ gives} \\ 1 &\le \frac{x}{\sin{x}} \le \frac{1}{\cos{x}} \\[12pt] 1 &\ge \frac{\sin{x}}{x} \ge \cos{x} \\ \end{align*}

And now we take the limit as \(x \to 0^+\) (i.e. \(x\) will move towards zero through positive numbers. Clearly \[ \lim_{x \to 0^+} 1 = \lim_{x \to 0^+} \cos{x} = 1\] because both are continuous functions that are defined at \(x=0\) and both take the value \(1\) at \(x = 0\). Thus \[ \lim_{x \to 0^+} 1 \ge \lim_{x \to 0^+} \frac{\sin{x}}{x} \ge \lim_{x \to 0^+} \cos{x} \\ 1 \ge \lim_{x \to 0^+} \frac{\sin{x}}{x} \ge 1 \\ \Longrightarrow \lim_{x \to 0^+} \frac{\sin{x}}{x} = 1 \]

All that remains is to deal with the case for negative angles. We do this by writing \(y=-x\) and then we have \begin{align*} \lim_{y \to 0^-} \frac{\sin{-y}}{-y} &= 1 \\ \lim_{y \to 0^-} \frac{-\sin{y}}{-y} &= 1 \\ \lim_{y \to 0^-} \frac{\sin{y}}{y} &= 1 \\[12pt] \left. \begin{array}{c} \lim_{x \to 0^+} \frac{\sin{x}}{x} = 1 \\ \lim_{x \to 0^-} \frac{\sin{x}}{x} = 1 \end{array} \right\rbrace \Longrightarrow \lim_{x \to 0} \frac{\sin{x}}{x} &= 1 \end{align*} as required.

Summing the Angles

I now want to draw another picture.

There, that's nice! A unit circle again, with lots of additions. What shall we do with it? Well, since it is the unit circle \(\overline{OP} = 1\) so \(\overline{BP} = \sin{(a+b)}\) and \(\overline{OB} = \cos{(a+b)}\). And \(\angle{QRP}\) being a right-angle, we have \(\overline{PR} = \sin{b}\) and \(\overline{OR} = \cos{b}\).

Now, \(\angle{ORS} = a\), giving \(\angle{PRS} = \frac{\pi}{2} - a\) and hence \(\angle{RPS} = a\) as marked. This means that \[\frac{\overline{PS}}{\overline{PR}} = \frac{\overline{OA}}{\overline{OR}} = \cos{a} \\ \Longrightarrow \overline{PS} = \sin{b} \cos{a} \qquad \overline{OA} = \cos{b} \cos{a}\] Finally, we have \(\overline{RS} = \overline{AB}\) and \(\overline{AR} = \overline{BS}\) \[ \frac{\overline{RS}}{\overline{PR}} = \frac{\overline{AR}}{\overline{OR}} = \sin{a} \\ \Longrightarrow \overline{AB} = \sin{b} \sin{a} \qquad \overline{BS} = \cos{b} \sin{a} \]

Putting all of this together we see that \begin{align*} \overline{BP} &= \overline{BS} + \overline{SP} \\ \Longrightarrow \sin{(a+b)} &= \sin{a} \cos{b} + \cos{a} \sin{b} \\[12pt] \overline{OB} &= \overline{OA} - \overline{AB} \\ \Longrightarrow \cos{(a+b)} &= \cos{a} \cos{b} - \sin{a} \sin{b} \\ \end{align*}

Now we will use these results (plus \(\sin -x = -sin x\) and \(\cos -x = cos x\)) to get two more. \begin{align*} \sin{(a+b)} - \sin{(a-b)} &= \sin{a} \cos{b} + \cos{a} \sin{b} - \sin{a} \cos{-b} - \cos{a} \sin{-b} \\ &= 2\cos{a} \sin{b} \\[12pt] \cos{(a+b)} - \cos{(a-b)} &= \cos{a} \cos{b} - \sin{a} \sin{b} - \cos{a} \cos{-b} + \sin{a} \sin{-b} \\ &= -2\sin{a} \sin{b} \end{align*}

Take It to the Limit One More Time

So we have derived two (or more) core results of trigonometry using little more than the definition of \(\sin{x} = \frac{o}{h}\) and \(\cos{x} = \frac{a}{h}\) where \(a\), \(o\) and \(h \require{cancel} \newcommand{\ddx}[1]{\frac{\mathbb{d}}{\mathbb{d}x}#1}\) are the adjacent, opposite and hypotenuse of a right-angled triangle.

Putting all these results together with the continuity of \(\cos\) and \(\sin\) will get us a couple more. \begin{align*} \ddx{\sin x} &= \lim_{k \to 0} \frac{ \sin{(x+k)} - \sin{x} }{k} \\ &= \lim_{k \to 0} \frac{ \sin{\left( (x+\frac{k}{2})+\frac{k}{2} \right)} - \sin{\left( (x+\frac{k}{2})-\frac{k}{2} \right)} }{k} \\ &= \lim_{k \to 0} \frac{ 2\cos{(x+\frac{k}{2})} \sin{\frac{k}{2}} }{k} \\ &= \lim_{k \to 0} \cos{(x+\frac{k}{2})} \cancelto{1}{\lim_{k \to 0} \frac{ \sin{\frac{k}{2}} }{\frac{k}{2}}} \\ &= \cos{x} \\ \end{align*} And finally \begin{align*} \ddx{\cos x} &= \lim_{k \to 0} \frac{ \cos{(x+k)} - \cos{x} }{k} \\ &= \lim_{k \to 0} \frac{ \cos{\left( (x+\frac{k}{2})+\frac{k}{2} \right)} - \cos{\left( (x+\frac{k}{2})-\frac{k}{2} \right)} }{k} \\ &= \lim_{k \to 0} \frac{ -2\sin{(x+\frac{k}{2})} \sin{\frac{k}{2}} }{k} \\ &= \lim_{k \to 0} -\sin{(x+\frac{k}{2})} \cancelto{1}{\lim_{k \to 0} \frac{ \sin{\frac{k}{2}} }{\frac{k}{2}}} \\ &= -\sin{x} \\ \end{align*}

A Parting Shot

One beautiful result I've recently come across is given below \[\lim_{x \to 0} \frac{1}{x^2}- \frac{1}{\sin^2 x}\]

In order to get there smoothly, I first want to establish the triple angle formula for \(\sin\). \begin{align*} \sin{3x} &= \sin{2x}\cos{x} + \cos{2x}\sin{x} &= 2\sin{x}\cos^2{x} + (1-2\sin^2{x})\sin{x} \\ &= 2\sin{x}(1-\sin^2{x}) + (1-2\sin^2{x})\sin{x} &= 2\sin{x} - 2\sin^3{x} + \sin{x} - 2\sin^3{x} \\ &= 3\sin{x} - 4\sin^3{x} \end{align*}

Now let's proceed \begin{align*} \lim_{x \to 0} \left( \frac{1}{x^2}- \frac{1}{\sin^2 x} \right) &= \lim_{x \to 0} \left( \frac{1}{x} - \frac{1}{\sin x} \right)\left( \frac{1}{x} + \frac{1}{\sin x} \right) \\ &= \lim_{x \to 0} \frac{1}{x}\left( \frac{1}{x} - \frac{1}{\sin x} \right)\left( 1 + \frac{x}{\sin x} \right) \\ &= \lim_{x \to 0} \left( \frac{1}{x^2} - \frac{1}{x\sin x} \right) \cancelto{2}{\lim_{x \to 0} \left( 1 + \frac{x}{\sin x} \right)} \\ &= 2\lim_{x \to 0} \frac{\sin x - x}{x^2\sin x} \\ &= 2\lim_{x \to 0} \left( \frac{\sin x - x}{x^3} \cdot \frac{x}{\sin x} \right) \\ &= 2\lim_{x \to 0} \frac{\sin x - x}{x^3} \cancelto{1}{\lim_{x \to 0} \frac{x}{\sin x}} \\ \end{align*} Now we suppose that \begin{align*} L = \lim_{x \to 0} \frac{\sin x - x}{x^3} &= \lim_{x \to 0} \frac{3\sin \tfrac{x}{3} - 4\sin^3 \tfrac{x}{3} - 3\tfrac{x}{3}}{27\left(\tfrac{x}{3}\right)^3} \\ &= \lim_{x \to 0} \frac{3\sin \tfrac{x}{3} - 3\tfrac{x}{3}}{27\left(\tfrac{x}{3}\right)^3} - \lim_{x \to 0} \frac{4\sin^3 \tfrac{x}{3}}{27\left(\tfrac{x}{3}\right)^3} \\ &= \tfrac19 \lim_{x \to 0} \frac{\sin \tfrac{x}{3} - \tfrac{x}{3}}{\left(\tfrac{x}{3}\right)^3} - \tfrac{4}{27} \left( \cancelto{1}{\lim_{x \to 0} \frac{\sin \tfrac{x}{3}}{\tfrac{x}{3}}} \right)^3 \\ \text{and writing $y = \tfrac{x}{3}$} \qquad &= \tfrac19 \lim_{y \to 0} \frac{\sin y - y}{y^3} - \tfrac{4}{27} \\ &= \tfrac19 L - \tfrac{4}{27} \\ \tfrac89 L &= -\tfrac{4}{27} \\ L &= -\tfrac16 \\ \end{align*} And so we have $$\lim_{x \to 0} \left( \frac{1}{x^2}- \frac{1}{\sin^2 x} \right) = 2\lim_{x \to 0} \frac{\sin x - x}{x^3} = -\tfrac13$$