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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.

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    Maybe you can write down the concrete example where you have difficulties, so we may help you. – 2017-02-09
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    Ok, I see you meant an expression of the form $0$ * $\infty$ where the zero factor is the actual constant $0$. Yes, you can do ordinary algebraic simplifications (as long as all terms are defined on the domain relevant to the limit), so in that case $0$ * $\infty$ can be replaced by zero before taking the limit. – 2017-02-09
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    As Momo suggested, please give a concrete example that you have concerns about. – 2017-02-09
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    If I see your question correct, you have, as part of your limit, $x \rightarrow \infty$ $\frac{e^x}{e^x}$. Although both the denominator and numerator do approach infinity, they approach it at the exact same rate (since they are the same function). That part of the limit just reduces to 1. – 2017-02-09
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    Same principle: $\frac{e^x}{e^x}$ is equal to $1$ for all x, so can be simplified away before taking the limit. No need to regard it as an indeterminate form. – 2017-02-09
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    In fact, even if that factor had been $\frac{e^x + 1}{e^x + 2}$, you could still evaluate the limit of that factor (which equals $1$ as $x$ approaches $\infty$), and hence that factor could be replaced by $1$ in the full expression whose limit you are trying to find. – 2017-02-09
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    If you think that "$\log(1+1/x)=(\log 1)\cdot(\log 1/x)$", i.e. that "$\log(A+B)=(\log A)\cdot(\log B)$", you need a **very serious** review of the properties of logarithmic functions. – 2017-02-09
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    Ross analysis is correct. If denominator tends to $\infty$ you don't need to bother about behavior of numerator. Apply the rule and if $f'(x)/g'(x)$ tends to a limit then you have solved the problem. – 2017-02-09

2 Answers 2

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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.

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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 $L\neq 0$.)

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    I removed a part of my answer which said l'Hôpital doesn't apply and gave an example, but it wasn't actually a counterexample, rather just an example where the limit of the fraction of derivatives doesn't exist while the original limit does. – 2017-02-09
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    @RRL: Thanks, but I like to think correcting my mistakes is its own reward. – 2017-02-09