# L'Hôpital's rule for vector-valued functions

What are the traps, when using L'Hôpital's rule in multivariable calculus to determine the lim of an vector function? I heard there some more cases, where L'Hôpital's rule in multivariable calculus is not applicable.

Update: I meant multivariable calculus. Its a bit difficult to translate math problems from german to english.

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I don't know what you mean by "dimensional analysis". Do you just mean multivariable calculus, or is this something else? – Gerry Myerson Apr 16 '13 at 12:49
@GerryMyerson, it's more of Physics Term. That is why you might not be familiar. See – user45099 Apr 16 '13 at 12:51
@user57, thanks. I'm vaguely familiar with that use of the term, but I don't see what it has to do with l'Hopital and limits of vector functions. Maybe someone will give a good answer, and I'll learn something. – Gerry Myerson Apr 16 '13 at 13:09

The proof of de l'Hôpital's rule is based on the mean value theorem (MVT) which guarantees the existence of a point $\xi\in\ ]a,b[\$ such that $$f(b)-f(a)=f'(\xi)(b-a)\ .$$ The MVT is usually proven by means of Rolle's theorem, and the latter is strictly about real-valued functions, insofar as the standard proof is based on the fact that a continuous function $f:\ [a,b]\to{\mathbb R}$ takes a maximum on $[a,b]$.
The example $$f(t):=e^{it}\qquad(0\leq t\leq2\pi)$$ shows that the MVT does not hold for complex- or vector-valued functions.
The point $\xi$ does not appear in de l'Hôpital's rule anymore. Therefore producing a counterexample is more tricky. Here is one: Let $$f(t):=t,\quad g(t):=t e^{-i/t}\ \qquad (t>0).$$ Then $\lim_{t\to0+} f(t)=\lim_{t\to0+} g(t)=0$; furthermore $$f'(t)\equiv 1,\quad g'(t)=\left(1+{i\over t}\right)e^{-i/t}\ .$$ It follows that $$\lim_{t\to0+}{f'(t)\over g'(t)}=\lim_{t\to0+}\left({t\over t+i} e^{i/t}\right)=0\ .$$ But the $\lim_{t\to0+}{f(t)\over g(t)}=\lim_{t\to0+}e^{i/t}$ does not exist.
There is a way out that also works in the multivariable case: Compute Taylor approximations to everything in sight, check for possible cancellations and hopefully arrive at a limit of the form $$\lim_{x\to 0}{f_1(x)\over g_1(x)}\ ,$$ where at least one of $f_1$, $g_1$ has a limit at $x=0$ which is $\ne0$, $\ne\infty$. (In the above counterexample no such Taylor approximation is possible.)