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Say we have a function $f:\mathbb{R}\to\mathbb{R}$, differentiable such that $\lim_{x\to\infty}f(x)=a, a\in\mathbb{R}$ and $\exists\lim_{x\to\infty}xf'(x).$ And I have to calculate: $$\lim_{x\to\infty}xf'(x)$$

...First let's rewrite this as:

$\lim_{x\to\infty}\frac{f'(x)}{\frac 1x}$. Now... Can I use "inverse l'hospital"? and what I mean by that is... to integrate the top and bottom and have: $\lim_{x\to\infty}\frac{f(x)}{ln(x)}$ and use the fact that $\lim_{x\to\infty}f(x)=a$.... so my limit would be $\lim_{x\to\infty}\frac {a + c_1}{ln(x)+c_2}=0?$

And if I can't do this, how can I solve this problem?

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  • $\begingroup$ i dont have a detailed proof why it wont work , but i can think of some examples where it wont work like in $\frac{\ln(x)}x, \frac{\sin(x)}x...$. $\endgroup$ – The Integrator Apr 23 '18 at 16:43
  • $\begingroup$ @pranavB23 Hmm... I see your point but do you think it might work in some cases? $\endgroup$ – C. Cristi Apr 23 '18 at 16:43
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    $\begingroup$ The anti-derivatives limit would have to be indeterminate (both 0, or both infinities) in order for the rule to apply to it. You'd need to prove that the anti-derivatives limit is indeterminate. Also you forgot about the constant of integration. $\endgroup$ – The Integrator Apr 23 '18 at 16:46
  • $\begingroup$ Reading backwards, you would apply the direct L'Hospital for $\frac{f'(x)}{\frac 1x}$ but this doesn't satisfy the criterium of the theorem, as $f'(x) $ does not tend to zero. $\endgroup$ – Berci Apr 23 '18 at 16:47
  • $\begingroup$ @pranavB23 I didn't forgot about the constant of integration but in this case it would still be 0 and yeah.. you're right the "primitive" limit should be an indeterminate $\endgroup$ – C. Cristi Apr 23 '18 at 16:48
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I am not convinced about the inverse of L'Hopital's rule at all. All I know is there are plenty of examples when it does not work.

Instead, you can use L'Hopital itself. The idea is that $(xf(x))' = f(x) + xf'(x)$.

Note that since $\lim_{x \to \infty} f(x) = a$ exists and $\lim_{x \to \infty} xf'(x)$ exists, we get that $\lim_{x \to \infty} (xf(x))'$ exists.

Now, all we need is the following transformation: $$ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{xf(x)}{x} = \lim_{x \to \infty} \frac{(xf(x))'}{1} = \lim_{x \to \infty} f(x) + \lim_{x \to \infty} xf'(x) $$

Where weare a slight generalization of L'Hopital(where the indeterminate form can be anything over infinity) by the fact that the denominator is infinite at infinity, and both numerator and denominator are at least differentiable in $x > R$ for some $R$, and that $\lim_{x \to \infty} (xf(x))'$ exists, since otherwise the inequality is not valid. That is why we need $\lim_{x \to \infty} xf'(x)$ to exist before knowing its value : L'Hopital cannot be used here otherwise.

From here, it is fairly clear that $\lim_{x\to \infty} xf'(x) = 0$. I will attach a link to the other L'Hopital rule shortly.

EDIT : In this answer :If $f'$ tends to a positive limit as $x$ approaches infinity, then $f$ approaches infinity you may find the generalized rule.

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    $\begingroup$ You should remark that if $a=0$, the calculation should be done for $f+1$ so L'Hopital's rule is valid. $\endgroup$ – Funktorality Apr 24 '18 at 5:41
  • $\begingroup$ You may see the modified version of L'Hopital's formula which I have given a link to. In particular, one does not require anything from the numerator, but rather only that the denominator goes to infinity, which is the case there. $\endgroup$ – астон вілла олоф мэллбэрг Apr 24 '18 at 6:34

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