Algebraically compute $\lim_{x \to 0}\frac{\sin x}{x}$ I know that: $$\lim_{x \to 0}\frac{\sin x}{x} = 1$$ 
However all proofs I have seen resort to some kind of geometric appeal or otherwise fall to L'Hôpital's rule or Taylor series. However, I want an to see an proof that is purely algebraic (I mean least geometric intuition, mainly not diagrams to convince the reader of something, but obvious algebraic statements one after another) that uses nothing advanced like series and [I mean accessible to ordinary calculus beginner]... I do not want want elipson delta proofs but something that has quite evidently, no gaps. 
Edit: Some comments have said sine cannot be defined without geometry or series. Well, I meant to say that we can assume that we know the unit circle definition. I only meant to say that we do not rely on diagrams in the proofs. We know the obvious facts that what sine  and radians are, etc. I did not mean it to take to formally and rigorously.  
Hope you guys have some elegant and delightful proof of the kind. It is something which is bothering me for quite a while. 

You might use the fundamental relation of trigonometry and the known limit $\lim_{x \rightarrow 0}\frac{1- \cos x}{x^2}= \frac{1}{2}$
$$\lim_{x \rightarrow 0} \frac{\sin x}{x}= \lim_{x \rightarrow 0} \frac{\sqrt{1-\cos^2 x}}{x}= \lim_{x \rightarrow 0} \sqrt{\frac{1-\cos^2x}{x^2}}= \lim_{x \rightarrow 0} \sqrt{\frac{(1-\cos x )}{x^2}(1+ \cos x)}= 1$$
