Simple calculus question: is applying l'Hopital's rule to $sin(x) / x$ really circular reasoning?

So I know this is a dumb question maybe, but it has been in my head recently. I once told someone, on the topic of introductory calculus, that using l'Hopital's rule to calculate $$\lim\limits_{x\rightarrow 0} \frac{\sin(x)}{x}$$ is circular reasoning, because the limit we are looking for is the definition of the derivative of $\sin(x)$ at $x=0$. They argued that it is not because you can work out the derivative of cosine via its series representation without ever recurring to the definition.

Now, there are two ways that I know of to obtain the series representation of cosine: the obvious one is Taylor expansion, which obviously requires knowledge of all the derivatives, the other one is to prove the following statement:

If there exists two functions $f$ and $g$ such that $f'=g$ and $g'=-f$, then they are unique.

Then you have to kind of get out of the hat the series representations and show that they respect the conditions listed above, which only involves deriving polynomials. Nonetheless, it seems to me that you still have to know the derivatives of sin and cos to then identify the two trigonometric functions with the series representations using their uniqueness.

My question is: is there really a way to obtain the series representation for sin and cos without knowing the derivatives, thus getting rid of the circularity of the reasoning?

This question is by no means practical and it's just a curiosity of mine, I'm not interested in other ways to calculate the limit, which is trivial.

EDIT:

I don't think this is a duplicate, as I said in my comment if you read the question that is linked, you will notice that what the OP means by "circular reasoning" is something on proving the limit using formulas about the area of circles, which has nothing to do with my question about series representations.

Answer

We make the following definitions: $$ \sin(x) = x - \frac{1}{3!}x^3+ \frac{1}{5!}x^5+ \cdots = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{2k+1} $$ and $$ \cos(x) = 1 - \frac{1}{2!}x^2+ \frac{1}{4!}x^4+ \cdots = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{2k} + \cdots$$ The series above converge for each $x \in \mathbb{R}$ by the ratio test. A well-known theorem of power series asserts that the derivative of a power series is indeed the power series of derivatives. This is not a trivial theorem, it requires some effort to prove. But, with the term-by-term differentiation theorem is obvious that \begin{align} \frac{d}{dx} \sin(x) &= 1+ \frac{2}{3!}x^2+ \cdots + \frac{2k+1}{(2k+1)!}x^{2k} + \cdots \\ &= 1+ \frac{1}{2!}x^2+ \cdots + \frac{1}{(2k)!}x^{2k} + \cdots \\ &= \cos(x) \end{align} Thus, applying L-Hopital's Rule: $$ \lim_{x \rightarrow 0} \frac{ \sin x}{x} = \lim_{ x \rightarrow 0} \frac{\cos(x)}{1} = \frac{\cos(0)}{1} = 1.$$ Notice, $\cos(0)=1$ and $\sin(0)=0$ are nearly manifest with the definitions I offer at the start of this answer.

Here's an interesting follow-up: how do we prove other aspects of the function theory of sine and cosine via series. For example, how to prove the $2\pi$-periodicity, or adding-angles formulas ? Much is known. I'll leave it at that for now.