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?

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 $\lim_{x\to a}\frac{f(x)}{g(x)}=d$. 
