# When does L'Hopital's rule pick up asymptotics?

I'm taking a graduate economics course this semester. One of the homework questions asks:

Let $$u(c,\theta) = \frac{c^{1-\theta}}{1-\theta}.$$ Show that $\lim_{\theta\to 1} u(c) = \ln(c)$. Hint: Use L'Hopital's rule.

Strictly speaking, one can't use L'Hopital's rule; at $\theta=1$, $u(c,\theta)$ is not an indeterminate form. However, if one naively uses it anyway, $$\lim_{\theta\to 1} \frac{c^{1-\theta}}{1-\theta} = \lim_{\theta\to 1}\frac{-\ln(c) c^{1-\theta}}{-1} = \ln(c).$$

More formally, using a change of variable $\vartheta = 1-\theta$ and expanding in a power series, \begin{align*} u(c,\theta) &= \frac{1}{\vartheta} \bigg( 1 + (\vartheta\ln(c)) + \frac{1}{2!}(\vartheta\ln(c))^2 +\frac{1}{3!}(\vartheta\ln(c))^3 + \cdots \bigg)\\ &= \frac{1}{\vartheta} + \ln(c) + \frac{1}{2!}\vartheta\ln(c)^2 + \frac{1}{3!}\vartheta^2\ln(c)^3 + \cdots \end{align*} has constant first order term $\ln(c)$.

Is it a coincidence that L'Hopital's rule is picking up this asymptotic term? More generally, when does a naive application of L'Hopital's rule pick up the asymptotic behavior of a function near a singularity?

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Actually, l'Hosptial applies, because the singularity is removable – AlexR Aug 26 '13 at 19:39
Wait... that isn't possible, the limit seems to diverge for $\theta$ near $1$? – Evan Aug 26 '13 at 19:40
@AlexR No, the singularity is a pole, not removable. – Neal Aug 26 '13 at 20:05
@Neal woha, you're correct. And that means, l'Hospital doesn't hold, and that means, the statement is false! O.o Thanks for the hint. So then there's nothing to answer - l'Hospital doesn't work. – AlexR Aug 26 '13 at 20:20

It seems very likely to me that there was a typo in the question, and that the intended limit was $\lim_{\theta\to0}u(c,\theta)$ with $$u(c,\theta) = \frac{c^{1-\theta}-1}{1-\theta}.$$ Or something of the like. Certainly, when a constant term necessary to induce the indeterminate form is forgotten in either the numerator or denominator, naïvely applying L'Hopital's rule produces the "correct" result.

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Good call. Does this phenomenon happen only when the numerator or denominator is off by a constant from producing an indeterminate form? – Neal Aug 26 '13 at 20:10
I never said that was the only time that occurs. My intuition is that with a little thought, a suitable counterexample can be made. Certainly, there are generally many examples in math where you can do the wrong thing and still end up with the right answer. – Omnomnomnom Aug 26 '13 at 20:35
Indeed you did not say that. If I come up with a counterexample I'll post it as an answer. In any case, your answer is almost certainly the correct interpretation of the problem, so I've accepted it. – Neal Aug 27 '13 at 0:57
Thank you. I had misread your comment and was unnecessarily indignant as a result, so sorry for that. Since you're looking for the right answer by the wrong means, I highly recommend the following as inspiration, if only for the laughs. – Omnomnomnom Aug 27 '13 at 1:00

One can imagine a situation in which L'Hospital's Rule does not apply, but gives the right answer. This is not one of them. The limit is not $\ln c$. A glance at the expression shows that the limit from the left is "$\infty$" and the limit from the right is "$-\infty$."

Remark: Suppose that for some constants $a,b,c,d$ $$\lim_{x\to a} \frac{f(x)-b}{g(x)-c}=d$$ and that limit can be calculated by L'Hospital's Rule. Then the Rule, wrongly applied, will report that $\lim_{x\to a}\frac{f(x)}{g(x)}=d$.

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Well yes, but you're not really answering my question: To first order, $u(c) = \frac{1}{1-\theta} + \ln(c)$. Is it coincidence that L'Hopital's rule is picking up the first-order behavior near the pole? – Neal Aug 26 '13 at 20:11
L'Hospital's Rule cheerfully ignores constants. So for example $\lim_{h\to 0}\frac{\cos h}{h}$ does not exist, but because $\lim_{h\to 0}\frac{\cos h-1}{h}$ does, L'Hospital's Rule reports the limit of that when you wrongly apply it to $\frac{\cos h}{h}$. – André Nicolas Aug 26 '13 at 20:24
It will equally happily report, when wrongly used, that $\lim_{h\to 0}\frac{\cos h-12}{h-99}=0$. – André Nicolas Aug 26 '13 at 20:28