# 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 $\vert\sin x\vert\lt \vert x\vert.$

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.

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] $\lim_{x\to 0}\cos x = 1.$