L'Hopital's rule in case of zero * inf / info Can L'Hopital's rule be applied to indeterminate form: $\frac{0 * \infty}{\infty}$?
Ross's Analysis seems to apply it to any case the denominator is $\infty$, while http://mathworld.wolfram.com/LHospitalsRule.html states that both the numerator and denominator must be so.
In general, if the numerator is $0 * \infty$, but the $0$ is not only in the limit, but even before the limit, can we just simplify the numerator to $0$? E.g. $log 1  + 1/x = log 1 * log 1/x$ is of form $0 * \infty$, but since the first factor is a constant $0$, it seems to me that this should just be $0$. Oops.

UPDATE
For example, while working on a problem involving limit as x approaches $\infty$ of $f(x) + f'(x)$, with f(x) and f'(x) having finite limits, the suggestion is to take $f(x) = f(x) * e^x / e^x$, which, applying L'Hopital's rule, goes to $(f(x) + f'(x))*e^x/e^x$.  I don't know how that's valid if limit of f(x) = 0.

Context
I will give the clear context and what is concerning me.  The problem is from Ross Analysis.  $f$ and $f'$ are defined and have finite limits as x approaches infinity.  Let lim as x approaches infinity of $f(x) + f'(x) = L$.  Show lim $f(x) = L$ and $f'(x) = 0$.  Note: I am using "lim" to mean limit as x approaches infinity
Ross's hint is: $f(x) = f(x)*e^x/e^x$.
I did as follows: lim $f(x)$ = lim $f(x) * e^x / e^x$, which is of form $\infty / \infty$ (at least if $f(x) \ne 0$).  Applying L'Hopital's rule, we get lim $f(x)$ = lim $f(x) + f'(x)$, which would indicate that lim $f'(x) = 0$, therefore lim $f(x) = L$, QED.  But I'm not convinced that this is correct if lim $f(x) = 0$, because then it's of the form I described above.
More generally, I would like to know if, for L'Hopital's rule, it's sufficient to show the denom. is inf, as Ross seems to state explicilty (Theorem 30.2), or the numerator must be inf as well (as stated by the Wolfram reference above).  I believe this is a well defined question; if it is not, please tell me what is missing.
 A: The asymptotic behavior of the numerator $h = fg$ is completely irrelevant as long as $g'(x) \neq 0$ eventually, $g(x) \to \infty$ and $h'(x)/g'(x)$ converges to a finite limit as $x \to \infty$.
Suppose $h'(x)/g'(x) \to L$, then given $\epsilon >0$ there exists $K$ such that $|h'(x)/g'(x) -L| < \epsilon/2$ for all $x > K$. Consider any $x > x_1 > K$.
By the mean value theorem, there exists $\xi$ between $x$ and $x_1$ such that
$$\frac{h(x) - h(x_1)}{g(x) - g(x_1)} = \frac{h'(\xi)}{g'(\xi)},$$
and
$$L - \epsilon/2 < \frac{h(x) - h(x_1)}{g(x) - g(x_1)} = \frac{h'(\xi)}{g'(\xi)} < L + \epsilon/2.$$
Hence,
$$L - \epsilon/2 < \frac{\frac{h(x)}{g(x)} - \frac{h(x_1)}{g(x)}}{1 - \frac{g(x_1)}{g(x)}} < L + \epsilon/2 \\ \implies L - \epsilon/2 +\frac{h(x_1)- (L - \epsilon/2)g(x_1)}{g(x)}< \frac{h(x)}{g(x)}< L + \epsilon/2 + \frac{h(x_1)- (L + \epsilon/2)g(x_1)}{g(x)}.$$
Since $g(x) \to \infty$ and $C/g(x) \to 0$ we have for all $x$ sufficiently large and $x_1 > K$ fixed,
$$-\frac{h(x_1)- (L - \epsilon/2)g(x_1)}{g(x)} > - \epsilon/2, \\ \frac{h(x_1)- (L + \epsilon/2)g(x_1)}{g(x)} < \epsilon/2. $$
Thus, for all $x$ sufficiently large 
$$L - \epsilon < \frac{h(x)}{g(x)} < L + \epsilon,$$
and $h(x)/g(x) \to L.$ Nothing about the behavior of $h(x)$ as $x \to \infty$ was needed.
A: Here is an alternative way to handle this that doesn't require expanding l'Hôpital's rule.  You can turn it into a $\dfrac00$ limit by taking $\dfrac{f\cdot\frac{1}{h}}{\frac1g}$ instead, so that you can apply l'Hôpital (if the corresponding limit of derivatives exists).  
In your particular context, this would mean using l'Hôpital on $\dfrac{f(x)e^{-x}}{e^{-x}}$, which works not only when $f\to0$ but for any finite limit $f\to L$, to show that $f-f'\to L$ and hence $f'\to 0$.
(Alternatively, you can get the case where $f\to 0$ from the case where $f\to 1$ by considering $f+1$, so you may assume without loss of generality that $L\neq 0$.)
