The proof of ddxsin(x) goes something like this:
lim
My doubt about this is that it uses the limit of \frac{\sin(h)}h and \frac{\cos(h)-1}{h} in the proof. However, proving these two limits uses L'Hopital's rule, which uses the derivative of \sin(x) and \cos(x) to prove the limits. This causes a circular argument because we're using the derivative of \sin(x) to prove the derivative of \sin(x). Is there a way to prove these two limits without using L'Hopital's rule or just looking at the graph, or is there a way to find \frac{d}{dx}\sin(x) without using these two limits?
There is a nice way to prove the limits using geometry here, but I'm wondering if there's a way to do it without using this, either.
Edit: The way I'm defining sine is by the unit circle definition, not the Taylor series one.
Answer
Here is a different approach that uses the integral definition of the arcsine function. We will deduce the limit of interest, \lim_{h\to 0}\frac{\sin(h)}{h}=1, without appeal to geometry or differential calculus. (Note that \cos(h)-1=-2\sin^2(h/2))
Instead, we only rely on elementary analysis of continuous functions and their inverses along with simple properties of the Riemann integral. To that end, we now proceed.
We define the sine function, \sin(x), as the inverse function of the function f(x) given by
\bbox[5px,border:2px solid #C0A000]{f(x)=\int_0^x \frac{1}{\sqrt{1-t^2}}\,dt }\tag 1
for |x|< 1.
NOTE: It can be shown that the sine function defined as the inverse of f(x) given in (1) has all of the familiar properties that characterize the circular function \sin(x).
It is straightforward to show that since \frac{1}{\sqrt{1-t^2}} is positive and continuous for t\in (-1,1), f(x) is continuous and strictly increasing for x\in (-1,1) with \displaystyle\lim_{x\to 0}f(x)=f(0)=0.
Therefore, since f is continuous and strictly increasing, its inverse function, \sin(x), exists and is also continuous and strictly increasing with \displaystyle \lim_{x\to 0}\sin(x)=\sin(0)=0.
From (1), we have the bounds (SEE HERE)
\bbox[5px,border:2px solid #C0A000]{1 \le \frac{f(x)}x\le \frac{1}{\sqrt{1-x^2}}} \tag 2
for x\in (-1,1), whence applying the squeeze theorem to (2) yields
\lim_{x\to 0}\frac{f(x)}{x}=1 \tag 3
Finally, let y=f(x) so that x=\sin(y). As x\to 0, y\to 0 and we can write (3) as
\lim_{y\to 0}\frac{y}{\sin(y)}=1
from which we have
\bbox[5px,border:2px solid #C0A000]{\lim_{y\to 0}\frac{\sin(y)}{y}=1}
as was to be shown!
NOTE:
We can deduce the following set of useful inequalities from (2). We let x=\sin(\theta) and restrict x so that x\in [0,1). In addition, we define new functions, \cos(\theta)=\sqrt{1-\sin^2(\theta)} and \tan(\theta)=\sin(\theta)/\cos(\theta).
Then, we have from (2)
\bbox[5px,border:2px solid #C0A000]{y\cos(y)\le \sin(y)\le y\le \tan(y)}
which are the familiar inequalities often introduced in an introductory geometry or trigonometry course.
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