Limit of sin x. Two questions about this proof

Question

I'm trying to learn the derivative of $\sin x = \cos x$ and I'm a bit confused about my textbook's proof in a couple of places. Here is the text:

I don't have much issue with this first area using the longform version of derivatives by using limits:

Here is the area that I have a few issues with:

1. Why does the line segments $EB = EB$ ? The text replaces EB with ED during the proof.

2. Why does the text say: the function $(sin \theta/\theta)$ is an even function, so its right and left limits must be equal. How do they know it's even? Does this mean that when $sin \theta$ is negative, so is $\theta$?

Answer 1

You have asked the following questions.

1) Why does ae=ed

The answer to this question is that they did not claim or used AE=ED at all. They are using AE+ED=AD which makes perfect sense. They used EB < ED which is also true.

2) Why does the text say: the function (sinθ/θ) is an even function, so its right and left limits must be equal. How do they know it's even? Does this mean that when sinθis negative, so is θ?

Note that $ sin(-\theta )= -sin(\theta)$ and $(-\theta )=-\theta $ so the quotient $ \frac {sin \theta }{\theta }$ does not change when you change your $\theta $ to -$\theta $

Thus $ \frac {sin \theta }{\theta }$ is an even function.

Answer 2

1) $EB^2 = OE^2-OB^2 =OE^2 - r^2 = OE^2 - OA^2 = EA^2.$

2) $$f(x)\equiv\frac{\sin x}{x} =\frac{-\sin x}{-x} = \frac{\sin (-x)}{-x} - f(-x)$$ so $f(x) = \frac{\sin x}{x}$ is an even function. The left and right limits of an even function at the point $x = 0$ must be equal. It is easiest if you sketch the graph:

An even function is a function whose graph is symmetric with respect to the $y$ axis and the point $x = 0$ is on the $y$ axis as well; therefore, any sequence approaching the point $x = 0$ from one side, must have a corresponding "negative" sequence approaching in the same manner and both coincide on the point $x = 0$ at the limit.