Is using L'Hospital's Rule to prove $\lim \limits_{x \to 0} \sin{x}/x$ circular? If  $\frac{d(\sin{x})}{dx}= \cos{x}$  is proven using the limit $\lim \limits_{x \to 0}\frac{\sin{x}}{x}=1$ (as it is in most textbooks), would it be circular to then use $\frac{d(\sin{x})}{dx}= {\cos{x}}$ and L'Hospital to prove the
limit $\lim \limits_{x \to 0}\frac{\sin{x}}{x}=1$ (because limit $\lim \limits_{x \to 0}\frac{\sin{x}}{x}= \frac{\cos{0}}{1})$ later in the same textbook?
 A: You can establish that the derivative of $\sin x$ is $\cos x$ without the fact that $\sin x = x +o(x)$. 
Let $ x$ be in radians and define $\cos x$ and $\sin x$ as the coordinates of the point $P(\cos x ,\sin x)$ in the unit circle that spans a circular sector of area $x/2$.
We can define the area of this sector in terms of $x$, as 
$$A(x) = \frac{x\sqrt{1-x^2}}{2}+\int\limits_x^1 \sqrt{1-t^2}dt$$
Then we define $\cos x$ as the unique real number in $[-1,1]$ such that
$$A(\cos x) = \frac{x}{2}$$
and $\sin x = \sqrt{1-\cos^2 x}$
We can see that for $-1\leq x\leq1$, $A(x)$ is differentiable, and we get
$$A'(x) = -\frac{1}{2\sqrt{1-x^2}}$$
But since $B(\cos x) = 2A(\cos x) = x$, we have that $B$ is the inverse of $\cos x$, so we have that
$$(\cos x)' = B^{-1}(x)'$$
$$(\cos x)' = \frac{1}{B'(B^{-1}(x))}$$
$$(\cos x)' = \frac{1}{\frac{-1}{\sqrt{1-B^{-1}(x)^2}}}$$
$$(\cos x)' = -\sqrt{1-\cos^2 x}$$
$$(\cos x)' = -\sin x$$
Since $\sin x = \sqrt{1-\cos^2 x}$, we have that
$$(\sin x)' = \frac{-2 \cos x (-\sin x)}{2\sqrt{1-\cos^2 x}}=\frac{\cos x \sin x}{\sin x} = \cos x$$
So, from my point of view, it isn't.
