TITLE: The Intuition behind l'Hopitals Rule
QUESTION [81 upvotes]: I understand perfectly well how to apply l'Hopital's rule, and how to prove it, but I've never grokked the theorem. Why is it that we should expect that $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x \to a}\frac{f'(x)}{g'(x)},$$
but only when specific criteria are met, like $f(x)$ and $g(x)$ approaching $0$ as $x \to a$? I've never found the proof of the theorem to shed much light on this (thought this might be because I don't have much of an intuitive sense as to why the Cauchy Mean Value Theorem holds, either). I've always been mystified as to how l'Hopital (Bernoulli?) ever discovered this theorem, beyond accidentally discovering that it holds in some cases and then generalizing it to a theorem. So, if you have any ideas as to where the theorem came from, and why it makes intuitive sense, I'd appreciate it if you shared them!

REPLY [11 votes]: This is Joe's answer and Michael's answer but with pictures! Also this answer is for intuition not rigour.
We start with functions $y=f(x)$ and $y=g(x)$ that are differentiable and have roots at $x=a$. Something like (maybe $f(x)$ blue and $g(x)$ red)

Now as Michael says, when we zoom in on differentiable functions they look like lines ---  near a point, functions are well-approximated by their tangents. How good is this approximation: doesn't matter here really!
So we draw the tangents:

Now $f(x)$ is light blue and $T_f$ is light blue. Also $g(x)$ is green and $T_g$ is red. At this point we could zoom in or we could follow Joe's argument. Instead I am going to translate everything to the origin! 
In terms of your limit you can do this via $z=x-a$:
$$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{z\rightarrow 0}\frac{f(z)}{g(z)}:$$

Now zoom in (This should be a $z$-axis)... and of course $T_f(z)=f'(a)z$ and $T_g(z)=g'(a)z$ (where the derivatives are respect to $x$):

Now you have:
$$\frac{f(x)}{g(x)}=\frac{f(z)}{g(z)}\approx \frac{f'(a)z}{g'(a)z}=\frac{f'(a)}{g'(a)},$$
near $z=0\Leftrightarrow x=a$.
Maybe the translation is unnecessary if we write $T_f(x)=f'(a)(x-a)$ and $T_g(x)=g'(a)(x-a)$.