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A simple form of l'Hôpital's rule looks like this: If $u$ and $v$ are functions with $u(0)=0$ and $v(0)=0$, the derivatives $\dot{v}(0)$ and $\dot{v}(0)$ are defined, and the derivative $\dot{v}(0)\ne 0$, then \begin{align*} \lim_{x\rightarrow 0} \frac{u}{v} &= \frac{\dot{u}(0)}{\dot{v}(0)} \qquad . \end{align*}

To me, the clearest way to arrive at this result uses a little nonstandard analysis: Since $u(0)=0$, and the derivative $d u/d x$ is defined at $0$, $u(d x)=d u$ is infinitesimal, and likewise for $v$. By the definition of the limit, the limit is the standard part of \begin{equation*} \frac{u}{v} = \frac{d u}{d v} = \frac{d u/d x}{d v/d x} \qquad , \end{equation*} where by assumption the numerator and denominator are both defined (and finite, because the derivative is defined in terms of the standard part). The standard part of a quotient like $p/q$ equals the quotient of the standard parts, provided that both $p$ and $q$ are finite (which we've established), and $q \ne 0$ (which is true by assumption). But the standard part of $d u/d x$ is the definition of the derivative $\dot{u}$, and likewise for $d v/d x$, so this establishes the result.

The generalizations to $x\rightarrow a$, where $a\ne 0$, and $x\rightarrow \infty$ are pretty trivial with the changes of variable $x\rightarrow x-a$ and $x\rightarrow 1/x$.

But there are a bunch of other cases of l'Hôpital's that seem to me to involve toxic doses of case-splitting. There are cases where you have to differentiate more than once, and cases where the indeterminate form is $\infty/\infty$ rather than $0/0$.

Is it possible to treat all of this in a unified way, possibly using ideas from projective geometry or inversions with respect to a circle in the complex plane?

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    The case where you differentiate more than once follows from the case where you differentiate once, by induction. I think the case $\infty / \infty$ follows by the $0/0$ case by letting $f(x)/g(x) = (1/g(x))/(1/f(x))$ and go back to a $0/0$ case. When you differentiate this new case, you get $$ \lim_{x \to 0} \frac{1/g(x)}{1/f(x)} = \lim_{x \to 0} \frac{ -g'(x)/g(x)^2 }{ -f'(x)/f(x)^2 } = \lim_{x \to 0} \frac{f(x)^2/g(x)^2}{f'(x)/g'(x)}. $$ I've tried to figure out the logic of it, but I'm a little tired (this is my second "tired" post today... gosh, no energy) maybe you can complete my arg.2012-01-18
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    @PatrickDaSilva: Thanks for your comment. I'm sure there is an inductive proof of then nth-derivative case, but I don't think the most obvious method of induction works. One of the hypotheses of l'Hôpital's rule is that $\dot{v}\ne 0$, which fails in the cases where you need multiple differentiation. I think an NSA-style approach works, since the Leibniz notation $d^2u/dx^2$ can be taken as essentially a literal division of two infinitesimal numbers. It's not so much that I can't find proofs in books or on the web, it's that there seems to be more need for case-splitting than I'd like.2012-01-18
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    @PatrickDaSilva: Your $(1/g)/(1/f)$ idea seems to work nicely. It requires lots of uses of $\lim(ab)=\lim a \lim b$ and $\lim(a/b)=\lim a/\lim b$, which may create some complication in the case where $\lim b=0$, or in the case where you want to use l'Hopital's rule to prove that a limit doesn't exist.2012-01-18
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    You're a fan of non-standard analysis huh? It sure does make proofs seem cleaner, but it takes a lot of work to understand *why* it works. I hope you've read stuff about it.2012-01-18
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    @Ben Crowell: **Danger!** It is not possible to use l'Hôpital's rule to prove that a limit does not exist!! See e.g. the example on page 4 of http://math.uga.edu/~pete/2400diffmisc.pdf.2012-01-18
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    @BenCrowell Maybe [this](http://math.stackexchange.com/questions/105256/another-form-of-the-lhospitals-rule) can help.2012-04-20

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