circular reasoning in proving $\frac{\sin x}x\to1,x\to0$

If by "circular" you mean "related to circle" then there is a proof of $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$ which is circular.

If by "circular" you mean "a reasoning which uses its final result (to be established) as one of the intermediate steps to achieve the final result" then there is a proof of $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$ which is not circular.

I assume we are using the second meaning of "circular" here. The standard proof via using area of sector of circle to derive the inequalities $\sin x < x < \tan x$ is pretty well known and described by moderator robjohn in this answer.

There is no inconsistency in the above proof if we understand that this proof relies on the following facts:

1) A point $P$ on the unit circle $x^{2} + y^{2} = 1$ has the coordinates $(\cos \theta, \sin \theta)$ by definition where $\theta/2$ is the area of sector $AOP$ where $A = (1, 0), O = (0, 0)$. Note that the area of sector is also used to define measure of an angle and thus we say that the measure of $\angle AOP$ is $\theta$ (twice the area of sector $AOP$).

2) Area of triangle $AOP$ is less than area of sector $AOP$ which is further less than area of triangle $OAB$ where $AB$ is tangent to unit circle at $A$ and $OPB$ is a straight line.

3) The fundamental constant $\pi$ can be defined as total area of the unit circle which gives rise to $\sin (\pi/2) = 1, \cos (\pi / 2) = 0$.

These facts require the existence of area of sector of circle which is justified by the existence of the integral $\int_{0}^{x}\sqrt{1 - t^{2}}\,dt$ for all $x \in [0, 1]$. The additive properties of area can also be established if we use the definition of area as an integral and this justifies the inequalities between areas mentioned in 2) above.

Note that the definitions of $\sin \theta, \cos \theta$ as mentioned in 1) are fully rigorous without any circular reasoning. Just to elaborate I will establish the formula $$\int_{0}^{x} \sqrt{1 - t^{2}}\,dt = \frac{x\sqrt{1 - x^{2}}}{2} + \frac{1}{2}\sin^{-1}x$$ directly from definitions of $\sin, \cos $ above.

For this purpose set point $P$ on unit circle with coordinates $(x, \sqrt{1 - x^{2}})$. Let $O = (0, 0)$ be origin $A = (x, 0), B = (0, 1)$ and integral above represents area of $OAPB$ where $PB$ is the arc of circle. This area is clearly sum of areas of triangle $OAP$ and sector $OPB$. Clearly area of triangle $OAP$ is $$\frac{1}{2}\cdot OA \cdot AP = \frac{x\sqrt{1 - x^{2}}}{2}$$ Now we need to show that area of sector $OPB$ is $(\sin^{-1}x)/2$.

Let $C = (1, 0)$ and note that if $\angle POC = \theta$ then $$\sin \theta = AP = \sqrt{1 - x^{2}}$$ and then $$\begin{aligned}\angle BOP &= 2\cdot \text{area of sector }BOP\\ &= 2(\text{area of sector }BOC - \text{ area of sector }POC)\\ &= 2\cdot\frac{\text{area of unit circle}}{4} - 2\cdot\text{area of sector }POC\\ &= \frac{\pi}{2} - \angle POC = \frac{\pi}{2} - \theta\end{aligned}$$ Thus we get $$\sin \angle BOP = \sin\left(\frac{\pi}{2} - \theta\right) = \cos \theta = \sqrt{1 - \sin^{2}\theta} = x$$ It follows that $$\text{area of sector }BOP = \frac{1}{2}\angle BOP = \frac{1}{2}\sin^{-1}x$$

The Archimedes method:

Take it to be a n-sided polygon, ther area of such n-gon will be: $$A=n\left(\frac12bh\right)$$ Rearranging: $$A=\frac12(nb)h$$ where actually $nb$ is perimeter of the n-gon, so: $$A=\frac12Ph$$ when we increase n to a very large value, h just becomes the radius of a circle and from, the definition of $\pi$: $$\pi=\frac{P_{circle}}{diameter}=\frac P{2h}$$ So in that case: $$A\approx\frac12h(2\pi h)=\pi h^2$$