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Let $f(x)=e^{-2x}(\cos x+2\sin x)$ and $g(x) = e^{-x}(\cos x+ \sin x).$ Find all the errors (if any) in the following L'Hôpital's rule argument:

$\lim\limits_{x\to \infty}\dfrac{f(x)}{g(x)}=\lim\limits_{x\to \infty}\dfrac{f'(x)}{g'(x)}=\lim\limits_{x\to \infty} \dfrac{5}{2}e^{-x}=0.$

Here's my work.

Recall the requirements for L'Hôpital's Rule:

To argue that $\lim\limits_{x\to c}\dfrac{f(x)}{g(x)}=\lim\limits_{x\to c}\dfrac{f'(x)}{g'(x)},$ the following must be true:

$1.$ $f(x)$ and $g(x)$ are differentiable on an open interval $I,$ but not necessarily at some point $c.$

$2.$ $\lim\limits_{x\to c}f(x)=\lim\limits_{x\to c}g(x)=0$ or $\pm \infty.$

$3.$ $g'(x)\neq 0\;\forall x\in I, x\neq c.$

$4.$ $\lim\limits_{x\to c}\dfrac{f'(x)}{g'(x)}$ exists.

We show that $f(x)$ and $g(x)$ are differentiable on $(-\infty, \infty).$ We have that $f'(x) = e^{-2x}(-2(\cos x+2\sin x) +(-\sin x+2\cos x))=-5e^{-2x}\sin x\;\forall x\in \mathbb{R}.$ Also, $g'(x)=e^{-x}(-(\cos x+\sin x)+(-\sin x+\cos x))= -2e^{-x}\sin x\;\forall x\in\mathbb{R}.$ Note that when $g(x)=0,\dfrac{f(x)}{g(x)}$ is undefined. This occurs when $\cos x + \sin x = 0\Rightarrow \tan x = -1\Rightarrow x = \dfrac{3\pi}{4}+2n\pi,n\in\mathbb{Z}.$ Let $x_0$ be such that $\tan x_0 = -1.$ We thus have that $f(x_0)=e^{-2x_0}(-\dfrac{\sqrt{2}}{2}+\sqrt{2})$ and $g(x_0)=0.$ Hence $\dfrac{f(x_0)}{g(x_0)}$ is indeterminate. Also, consider when $x_1= \tan^{-1} \left(-\dfrac{1}{2}\right)+2n\pi.$ Then $\dfrac{f(x_1)}{g(x_1)}=\dfrac{e^{-2x_1}\left(\cos \left(\tan^{-1}\left(\dfrac{1}{2}\right)\right)-2\sin \left(\tan^{-1}\left(\dfrac{1}{2}\right)\right)\right)}{e^{-x_1}[\cos (\tan^{-1} (\frac{1}{2}))-\sin (\tan^{-1}(\frac{1}{2}))]}\\ =e^{-x_1}\dfrac{\frac{2}{\sqrt{5}}-\frac{2}{\sqrt{5}}}{\frac{2}{\sqrt{5}}-\frac{1}{\sqrt{5}}}=0.$

Hence $\dfrac{f(x)}{g(x)}$ is not indeterminate for all $x\in\mathbb{N}$ such that $x=\tan^{-1} (-\dfrac{1}{2})+2n\pi.$

Now consider $g'(x)=-2e^{-x}\sin x.$ $g'(x)=0$ whenever $\sin x=0$ as $e^{-x}\neq 0\;\forall x\in \mathbb{R}.$ Thus, $g'(x)=0\Leftrightarrow x=n\pi,n\in\mathbb{N}.$ So this is another error.

From above, we have that $\lim\limits_{x\to \infty}\dfrac{f'(x)}{g'(x)}$ does not exist since it is undefined whenever $x=n\pi,n\in\mathbb{N}$ and equal to $\lim\limits_{x\to \infty}\dfrac{-5e^{-2x}\sin x}{-2e^{-x}\sin x}=\lim\limits_{x\to \infty}\dfrac{5}{2}e^{-x}=0$ whenever $x\neq n\pi.$

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    "3) $g'(x)\neq 0;\forall x\in I, x\neq c$" and "$g'(x)=e^{-x}(-(\cos x+\sin x)+(-\sin x+\cos x))= -2e^{-x}\cos x;\forall x\in\mathbb{R}$ means failure doesn't it? $g'(n\pi + \frac {\pi}2) = 0$, right? – fleablood Nov 08 '19 at 23:49
  • yes it does. It must satisfy that $g'(x)\neq 0\forall x\in I$ but not necessarily at $c.$ Also, I made a careless mistake when I recalculated $g'(x).$ I fixed it though. –  Nov 08 '19 at 23:59
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    As you have demonstrated correctly the key issue here is the vanishing of $g'$ infinitely often as $x\to\infty $. You may have a look at related answer: https://math.stackexchange.com/a/1798950/72031 – Paramanand Singh Nov 09 '19 at 02:54
  • @ParamanandSingh Why the vanishing of g’ should be an issue. Isn’t the key issue the wrong application of HR at the first step. The second step alone should be correct since it leads to a correct evaluation for the limit without the application of HR. – user Nov 09 '19 at 03:02
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    @user : the proof of L'Hospital's Rule uses Cauchy's Mean Value Theorem and it requires the non-vanishing of derivative of denominator. If that's not guaranteed then the proof doesn't work. To put it more simply if the limit of $f'/g'$ exists and domains are intervals then $g'$ must not vanish on that interval. – Paramanand Singh Nov 09 '19 at 03:04
  • @user : in general if one can avoid L'Hospital's Rule then one should just because it requires so many subtle hypotheses to check. – Paramanand Singh Nov 09 '19 at 03:05
  • @ParamanandSingh But here if I’m not wrong we are excluding l’Hopital since $f/g$ is not defined in an interval. Therefore the step which leads to $f’/g’$ is wrong and meaningless. Anyway the expression $f’/g’$ has limit zero even if $g’=0$ for “some” point. – user Nov 09 '19 at 03:08
  • Well here checking of $g$ vanishing is easier (or perhaps simpler) than vanishing of $g'$ and you don't need to figure $g'$ for that. Sometimes it may be the case that $g'$ is a simpler expression and then vanishing of $g'$ also renders application of L'Hospital's Rule invalid. – Paramanand Singh Nov 09 '19 at 03:11
  • @user: you should have a look at my linked answer in the comment. It shows how the non vanishing of $g'$ also implies non-vanishing of $g$ in the $0/0$ case of L'Hospital's Rule. – Paramanand Singh Nov 09 '19 at 03:13
  • +1 for asking a nice question with lot of subtleties. – Paramanand Singh Nov 09 '19 at 03:14

2 Answers2

2

Hint What is $\frac{f(x)}{g(x)}$ when $$x= \frac{3\pi}{4}+2n \pi \,?$$

N. S.
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  • @FredJefferson That's one, which already implies that you cannot apply L'H. There is a second different condition in L'H which is not satisfied: you need $g'(x) \neq 0$, which also fails. – N. S. Nov 08 '19 at 23:08
  • And technically, the equality $\lim\limits_{x\to\infty} \dfrac{-5e^{-2x}\sin x}{-2e^{-x}\sin x}=\lim\limits_{x\to \infty}\dfrac{5}{2}e^{-x}$ is false, the LHS is not defined at $n \pi$ so the LHS limit does not exist. – N. S. Nov 08 '19 at 23:10
  • @FredJefferson Which means that $\lim \frac{f'(x)}{g'(x)}$ doesn't exist, so you cannot apply L'H... But this is a consequence of the fact that $g'(x)=0$ infinitely many times, so I am not sure if I would call it a mistake. – N. S. Nov 08 '19 at 23:12
  • @N.S. I'm not sure that the step $$\lim\limits_{x\to\infty} \dfrac{-5e^{-2x}\sin x}{-2e^{-x}\sin x}=\lim\limits_{x\to \infty}\dfrac{5}{2}e^{-x}$$ as a step alone is wrong. – user Nov 08 '19 at 23:15
  • @user Whenever when we write $\lim_{x \to \infty} h(x)$ we understand that there exists a $c$ such that $h(x)$ is defined for $x >c$. The limit on the left does not exist. The limit on the right is 0. They cannot be equal. – N. S. Nov 08 '19 at 23:18
  • @N.S. According to the more general definition of limit we should exclude the points such that $\sin x =0$ therefore, in that context, the two limits are equivalent. – user Nov 08 '19 at 23:22
  • @user there must only be one solution. I think that the definition of a limit as $x\to \infty$ is that $\forall \epsilon >0,\forall y\in\mathbb{R},\exists \delta >0$ such that $|x-y|<\delta \Rightarrow |f(x)-L|<\epsilon,$ where $L$ is the limit. –  Nov 08 '19 at 23:34
  • @N.S. I don't think that your answer is the point here. The points where the expression is not well defined should be excluded by the domain. For example what is the limit $$\lim_{x\to 0} \frac{\sin \frac 1x}{\sin \frac 1x}$$ It is one of course. – user Nov 08 '19 at 23:40
  • @user No, it is not. If you start doing that, the limit laws do not work anymore. – N. S. Nov 08 '19 at 23:47
  • @N.S. This topic was discussed in the following OP. – user Nov 08 '19 at 23:48
  • @user Let $f(x)=\frac{x}{1-1_{\mathbb Q}(x)}$. Let $g(x)=\frac{x}{1_{\mathbb Q}(x)}$. What is $\lim_{x \to 0} f(x)$? What about $\lim_{x \to 0}g(x)$? – N. S. Nov 08 '19 at 23:49
  • @user $1_{\mathbb Q}(x)$ is 1 when $x$ is rational and 0 when $x$ is irrational. – N. S. Nov 09 '19 at 00:07
  • @user NO it cannot. The proof of the L'H rule requires $\frac{f(x)}{g(x)}$ to be defined, you cannot exclude points and apply L'H – N. S. Nov 09 '19 at 01:00
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    @user : the application of L'Hospital's Rule involves that the domain of definition of $f, g, f/g$ be intervals. This is not usually explicitly stated but the proof of L'Hospital's Rule uses intervals. – Paramanand Singh Nov 09 '19 at 02:23
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    @user: the concept of limit has a definition for functions defined on domains other than intervals so the expression in question has a limit ($0$) but L'Hospital's Rule can't be used here. – Paramanand Singh Nov 09 '19 at 02:24
  • @ParamanandSingh Are you sure that the limit is zero? I've found that the limit doesn't exist. My idea to exclude l'Hospital was grounded on the fact that the limitboth for $f(x)$ and $g(x)$ don't exists. I'll check better your claim about the limitation for the application on interval. Thanks! – user Nov 09 '19 at 02:29
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    @user: check any proof of L'Hospital's Rule and you will be convinced that it requires that domains are intervals. Further $f, g$ have domains as intervals here and they tend to $0$ also. It is the ratio $f/g$ whose domain is not an interval. Also the claim about limit of expression being $0$ is wrong. The trigonometric part behavior does cancel the vanishing nature of exponentials used. A detailed analysis like the one in your answer demonstrates that. Sorry on that front. – Paramanand Singh Nov 09 '19 at 02:34
  • @ParamanandSingh Thanks for your contribution, I’ll take a closer look to that tomorrow. Bye – user Nov 09 '19 at 02:38
  • @ParamanandSingh Anyway of course on f and g I’m clearly wrong and the limit is zero of course. I didn’t realize that the exponential term tends to zero...so it is trivial. That drive me in a wrong direction. – user Nov 09 '19 at 02:47
  • @user: I liked your analysis of the expression when $x=2\pi n+e^{-2\pi n} - (\pi/4)$. That was pretty smart and non-obvious. +1 for your answer. – Paramanand Singh Nov 09 '19 at 02:49
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We have that

$$\frac{e^{-2x}(\cos x+2\sin x)}{e^{-x}(\cos x+ \sin x)}=e^{-x}\frac{\cos x+2\sin x}{\cos x+ \sin x}=e^{-x}+e^{-x}\frac{\sin x}{\sqrt 2 \sin\left(x+\frac \pi 4\right)}$$

and for $x=n\frac \pi 2-\frac \pi4$

$$e^{-x}\frac{\sin x}{\sqrt 2 \sin\left(x+\frac \pi 4\right)} \to 0$$

but for $x=2\pi n+e^{-2\pi n}-\frac \pi4$

$$e^{\left(-2\pi n-e^{-2\pi n}+\frac \pi4\right)}\frac{\sin \left(e^{-2 \pi n}-\frac \pi4\right)}{\sqrt 2 \sin\left(e^{-2 \pi n}\right)}=$$

$$= e^{-e^{-2 \pi n}}\cdot e^{\frac \pi 4} \cdot \frac{\sin \left(e^{-n}-\frac \pi4\right)}{\sqrt 2}\frac{e^{-2 \pi n}}{\sin\left(e^{-2 \pi n}\right)} \to 1 \cdot e^{\frac \pi 4}\cdot \left(-\frac 12\right)\cdot 1=-\frac{e^{\frac \pi 4}}{2}$$

therefore the limit doesn't exist.

With reference to your question, according to the observation by Paramanand Singh, the only relevant issue here is that we can't apply l'Hospital at the first step since $f/g$ is not defined on an interval.

Ineed, note that according to the more general definition of limit

$$\lim\limits_{x\to \infty}\dfrac{f'(x)}{g'(x)}=\lim\limits_{x\to \infty}\dfrac{-5e^{-2x}\sin x}{-2e^{-x}\sin x}=\lim\limits_{x\to \infty}\dfrac{5}{2}e^{-x}=0$$

since we can take the limit excluding from the domain the points such that $\sin x =0$.

Refer to the related

user
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  • is my proof that it's not in an indeterminate form correct? –  Nov 09 '19 at 00:03
  • @FredJefferson Note also, as an aside note, that l'Hospital is also applicable for the case $f(x) \to a\neq 0$ and $|g(x)| \to \infty$. Refer to general proof. – user Nov 09 '19 at 00:26