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In my textbook there is an exercise about L'Hospital's Rule (LHR for short):

If $f(x)\in\ C^1 (a,+\infty)$,$\displaystyle\lim_{x\to+\infty}[f(x)+f'(x)]=k\in\mathbb R$, prove :$\displaystyle\lim_{x\to+\infty}f(x)=k$
Proof: $$\lim_{x\to+\infty}f(x)=\lim_{x\to+\infty}\frac{e^xf(x)}{e^x}=\lim_{x\to+\infty}\frac{e^xf(x)+e^xf'(x)}{e^x}=\lim_{x\to+\infty}[f(x)+f'(x)]=k$$

Stunning. Isn't it? It's just magical to come up with $1=\frac{e^x}{e^x}$ here. And inspired by this magic, a thought popped up into my mind. I came to ponder on this: will it also be true to say the following?

If $f(x)\in\ C^1 (a,+\infty)$,$\displaystyle\lim_{x\to+\infty}[f(x)-f'(x)]=k\in\mathbb R$, then $\displaystyle\lim_{x\to+\infty}f(x)=k$

Naturally I tried $1=\frac{e^{-x}}{e^{-x}}$ this time. Then I got this $$\lim_{x\to+\infty}f(x)=\lim_{x\to+\infty}\frac{e^{-x}f(x)}{e^{-x}}$$ Now if I'm allowed to apply LHR then I'm done. But nobody tells me it is a $\frac00$ pattern. Yet I still don't want to give up on this thought. I think maybe the condition $\displaystyle\lim_{x\to+\infty}[f(x)-f'(x)]=k\in\mathbb R$ will make sure that LHR is appliable here. (In fact I guess $f(x)$ might even bounded here. Just a guess, though, based on nothing reliable) But I get stuck and cannot go any further to prove or disprove the appliability of LHR here. Smart people in this site, please give me some help. Thanks in advance!


SPECIAL EDIT: Clarification about the LHR
I'm making this edit about LHR because some of you believe that LHR is appliable only for the $\frac00$ and $\frac{\infty}{\infty}$ patterns. I'm not being rude but it seems like a major misbelief. I have to clarify here that LHR also works for the $\frac{*}{\infty}$ pattern where $*$ is not necessarily infinity.
Here goes one rigorous statement of LHR:

L'Hospital's Rule
Suppose $f(x)$ and $g(x)$ are differentialbe on $(a,a+d]$ (where $d>0$), and $g'(x)\ne 0$. If $$\lim_{x\to a^+}f(x)=\lim_{x\to a^+}g(x)=0$$ or $$\lim_{x\to a^+} g(x)=\infty,$$ and $$\exists \lim_{x\to a^+} \frac{f'(x)}{g'(x)}\in \mathbb R^{*}$$ where $\mathbb R^{*}$ denotes the extended real line, then $$\lim_{x\to a^+}\frac{f(x)}{g(x)}=\lim_{x\to a^+}\frac{f'(x)}{g'(x)}$$

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A disproof is $f(x)=e^x$. $f(x)-f'(x)=0,$ but $f(x)$ has no limit

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  • $\begingroup$ I really don't get how OP could miss this counterexample $\endgroup$ – AnalysisStudent0414 Mar 20 '15 at 18:22
  • $\begingroup$ Oh yes... I should have realized $e^x$'s invariablity under differenciation.. thank you $\endgroup$ – Vim Mar 20 '15 at 18:22
  • $\begingroup$ @AnalysisStudent0414. Sorry I was stupid. Maybe it's because I was too anxious to prove that but thought little about disproving that :) $\endgroup$ – Vim Mar 20 '15 at 18:27
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    $\begingroup$ @AnalysisStudent0414. My impression was that the switch from a plus to a minus usually wouldn't make so much difference. But this time it surely teaches me. $\endgroup$ – Vim Mar 20 '15 at 18:29
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Does the second equality in your proof (for the exercise, not for your conjecture) work for $f(x) = e^{-x}$? Or perhaps I should say "Why does the second equality NOT work?" :)

BTW: I really like this idea -- it is, as you observe, really clever!

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  • $\begingroup$ I have checked but I think it does work for $e^{-x}$. $\endgroup$ – Vim Mar 20 '15 at 18:26
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    $\begingroup$ What you apply LHR, both numerator and denominator need to go to 0 or $\infty$. When $f(x) = e^-x$, the numerator is $e^0 = 1$, but the denominator, $e^x$ goes to infinity. So applying LHR is not allowed. (The conclusion of the alleged theorem still holds...but this argument doesn't prove the theorem.) $\endgroup$ – John Hughes Mar 20 '15 at 18:28
  • $\begingroup$ I'm sorry but I think I have to point out that when the denominator goes to infinity, the numerator does NOT necessarily go to infinity. ;) $\endgroup$ – Vim Mar 20 '15 at 18:31
  • $\begingroup$ If you want to apply LHR to $\lim_{x \to a} \frac{u(x)}{v(x)}$, it's essential that either (i) $\lim_{x \to a} u(x) = \lim_{x \to a} v(x) = 0$ or (ii) $\lim_{x \to a} u(x) = \lim_{x \to a} v(x) = \pm \infty$; otherwise you cannot conclude that $\lim_{x \to a} \frac{u(x)}{v(x)} = \lim_{x \to a} \frac{u'(x)}{v'(x)}$. In the example I gave ($f(x) = e^{-x}$), we have $u(x) = 1$ and $v(x) = e^x$; the hypotheses do not hold, so you cannot apply LHR. $\endgroup$ – John Hughes Mar 20 '15 at 18:55
  • $\begingroup$ Well, strictly speaking, your question asks for the “why” of a claim that is wrong. The equality holds for $e^{-x}$, since $0=0$. You just can't use LHR to derive or prove it. $\endgroup$ – Christopher Creutzig Mar 20 '15 at 21:20
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Let $f(x) = e^x+k$. Then $f(x)-f'(x) = k$.

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