calculus - Is the proof of the derivative of $sin(x)$ circular?

The proof of $\frac{d}{dx}\sin(x)$ goes something like this:

$$\begin{aligned} \lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h}=\lim_{h\to0}\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h}\\ =\lim_{h\to0}\frac{\sin(x)(\cos(h)-1)+\cos(x)\sin(h)}{h}\\ =\lim_{h\to0}\frac{\sin(x)(\cos(h)-1)}{h}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}\\ =0+\cos(x)\times1\\ =\cos(x) \end{aligned}$$

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.

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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 $\displaystyle \lim_{x\to 0}\sin(x)=\sin(0)=0$.

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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

$$\lim_{x\to 0}\frac{f(x)}{x}=1 \tag 3$$

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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!

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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.