A general rule for L'Hôpital's Rule In my text there are as many as $11$ theorems dealings with the different cases of L'Hôpital's Rule and due to the quantitative heaviness I often forget rules during problem session. Isn't there a general rule for L'Hôpital's Rule which is easy to remind so that I won't need to spend useless time recalling result?
 A: A more general approach would be where 
$$
\lim_{n\to \infty} \frac{f(x)}{g(x)} = \lim_{n\to \infty} \frac{\frac{d}{dx} f(x)}{\frac{d}{dx} g(x)} 
$$
Or to 0 of course
$$
\lim_{n\to 0} \frac{f(x)}{g(x)} = \lim_{n\to 0} \frac{\frac{d}{dx} f(x)}{\frac{d}{dx} g(x)} 
$$
Only when $\lim_{n\to \infty/0} f(x) = \lim_{n\to \infty/0} g(x)$ tend towards the same thing ($\infty_{+/-}, 0$)
A good example being
$$
\lim_{x->0} \frac{sin(x)}{x} = \lim_{x\to 0} \frac{cos(x)}{1} = 1
$$
A: I think of it as a sort of "algorithm". Let's say you want to calculate the limit of
$$\lim _{x \to \infty} \frac{f(x)}{g(x)}$$
If this limit tends to either $\frac{0}{0}$ or $\pm\frac{\infty}{\infty}$, then we apply the rule and differentiate $f$ and $g$ on their own. By this, I mean you don't use the quotient rule, but you just do $\frac{f'(x)}{g'(x)}$. Now, you calculate the limit of this. Again, if you get the same thing, you differentiate again.
Let's consider an example:

Evaluate
$$\lim_{x \to 0} {\frac{2\sin (x) -\sin (2x)}{x-\sin (x)}}.$$

Here, we have $f(x) = 2 \sin(x) - \sin(2x)$ and $g(x) = x - \sin (x)$. Straight away, you notice that as $x \to 0$, we get the fraction tending to $\frac{0}{0}$. So we apply L'Hopitals rule:
$$f'(x) = 2\cos (x) -2\cos (2x)$$
$$g'(x) = 1-\cos (x)$$
So now, we are calculating
$$\lim_{x\to 0}{\frac{2\cos (x) -2\cos (2x)}{1-\cos (x)}}$$
again, this goes to
$$\frac{2 - 2}{1 - 1} \to \frac{0}{0}$$
and so we apply L'Hopitals again. Now we have:
$$f''(x) = -2\sin (x) +4\sin (2x)$$
$$g''(x) = \sin (x)$$
Now, calulating the limit gives
$$ \lim_{x\to 0}{\frac{-2\sin (x) +4\sin (2x)}{\sin (x)}} \to \frac{0}{0}$$
and so AGAIN, we apply the rule. Differetiating the third time gives us
$$f'''(x) = -2\cos (x) +8\cos (2x)$$
$$g'''(x) = \cos (x)$$
So now taking the limit gives
$$ \lim_{x\to 0}{\frac{-2\cos (x) +8\cos (2x)}{\cos (x)}} \to {\frac{-2 +8}{1}} \to 6.$$
