Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

2 Answers 2

up vote 5 down vote accepted

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.

share|improve this answer
    
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
1  
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 $\lim_{x\to a}\frac{f(x)}{g(x)}=d$.

share|improve this answer
    
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
1  
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.