What is the difference between Cauchy's Rule and L'Hôpital's Rule? My teacher didn't mention there was a difference, but I think the two rules are not exactly equal to one another.
What is the difference between the two rules?
Is it the functions that intervene in the limit?
Thank you!
Edit: Apparently in my country there is a poor practice of calling L'Hôpital's Rule, the Cauchy's Rule.
Edit2: https://answers.yahoo.com/question/index?qid=20120131110851AAlLBcp
 A: You are probably thinking about Cauchy's theorem and l'Hôpital's rule. The difference is that Cauchy's theorem is used to prove the second one.

Cauchy's theorem. Suppose the functions $f(x)$ and $\varphi(x)$ are continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b).$ For simplicity, assume $$\varphi'(x)\neq0\ \ \forall x\in(a,b).$$ When these conditions are met, there exists a $c\in(a,b)$ such that $$\frac{f(b)-f(a)}{\varphi(b)-\varphi(a)}=\frac{f'(c)}{\varphi'(c)}.$$

As mentioned, this fact is applied in the proving of l'Hôpital's rule. The latter simplifies many limit problems where indeterminate forms are encountered. The famous limit $$\lim_{x\to0}\frac{\sin x}{x}$$ becomes straight-forward since
$$\lim_{x\to0}\frac{\sin x}{x}=\left\{\frac{0}{0}\right\}=\lim_{x\to0}\frac{(\sin x)'}{x'}=\lim_{x\to0}\cos x=1. \ \ (*)$$
Also, 
$$\lim_{x\to \infty}\frac{-x^3}{e^x\ln x}=\left\{\frac{\infty}{\infty}\right\}=-1\cdot\lim_{x\to \infty}\frac{(x^3)'}{(e^x\ln x)'}=0.$$
Cauchy's theorem, in turn, is shown to be true via Rolle's theorem.

(*) Here the trigonometric functions are defined as infinite series in order to avoid circular argumentation. (See comment section below.)
