Appendix - Limits of Special Functions

There are many special functions that have very nice limit properties. These are not necessarily easy to prove. We'll have a look at some limits of special functions here in this appendix. You can skip this if you believe the results.

We'll need to establish a couple of useful inequalities before we start cranking out results. It would be nice to use simple trigonometric identities for these, but at first we need to relate the angle \(x\) to trigonometric quantities so we can compare them. This requires a picture. Figure 1 shows a few triangles associated with the angle \(x\), superimposed on the unit circle. The length of the red segment is \(\sin x\). The length of the arc of the unit circle from A to D is \(x\) - the angle in radians. Observe that the red segment is perpendicular to OA, so point A is farther from point D than the point E is, so the length of the arc from A to D is greater than the length of the red segment. This immediately shows one very useful result:

Theorem: [SineCompare] If \(0\lt x\lt \frac\pi2\) then \[\sin x \lt x.\]
Note that if we basically turn the picture upside down, we also have a result for negative \(x\). In general we can write \[\vert\sin x\vert\lt \vert x\vert.\]

x 1 A B x C D E O
Figure 1: Triangles in the unit circle associated with angle \(x\)

The length of the blue segment is \(\tan x\). The triangle BCD is a right triangle whose hypotenuse is the segment from B to C. This means that the distance from B to C is greater than the distance from B to D. Thus the the path from A to B to D is shorter than the blue segment. On the other hand, the piecewise linear path from A to D passing through B is longer than the arc from A to D, since the arc lies inside the other path. This all means that

Theorem: [TanCompare] If \(0\lt x\lt \frac\pi2\) then \[x\lt \tan x.\]
Again we can tip the picture to get a result for negative \(x\) as well. We conclude through these simple geometric arguments that for any nonzero angle \(x\) between \(-\pi/2\) and \(\pi/2\) we have \[\vert\sin x\vert \lt \vert x\vert \lt \vert\tan x\vert.\]

Next, we want to find an inequality involving the cosine function. In order to avoid cluttering our picture excessively, we have reproduced it in Figure 2, with different segments emphasized.

x 1 A B x D E O
Figure 2: Triangles in the unit circle associated with angle \(x\)

We see that the length of the segment OE is \(\cos x\), so that the length of the cyan segment is \(1=\cos x\). The crimson segment is the base of a right triangle with hypotenuse AD. That hypotenuse lies inside the arc AD, which we already noted had length \(x\). This gives

Theorem: [CosCompare] If \(0\lt x\lt \frac\pi2\) then \[1-\cos x \lt x.\]
Another way of writing this is to say \[1-x\lt \cos x.\] When \(x\) is negative, we can tip the picture to show that \(1-\vert x\vert \lt \cos x.\) for all nonzero values of \(x\) between \(-\frac\pi2\) and \(\frac\pi2\).

At this point we can talk about limits of the basic trigonometric functions. For any positive \(\epsilon\) we can choose \(\delta=\epsilon\) and use [SineCompare] to see that \begin{align*} \vert \sin x - 0\vert \lt \vert x - 0\vert\lt \epsilon. \end{align*} This means that

Theorem: [SineZeroLimit] \[\lim_{x\to 0} \sin x = 0.\]
Likewise [CosCompare] shows that \begin{align*} \vert \cos x - 1\vert &\lt \vert x\vert\\ &\lt \epsilon, \end{align*} which establishes the following theorem.
Theorem: [CosineZeroLimit] \[\lim_{x\to 0}\cos x = 1.\]