5
$\begingroup$

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 $x\neq n\pi.$

$\endgroup$
  • 1
    $\begingroup$ "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? $\endgroup$ – fleablood Nov 8 at 23:49
  • $\begingroup$ 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. $\endgroup$ – user718615 Nov 8 at 23:59
  • 1
    $\begingroup$ 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: math.stackexchange.com/a/1798950/72031 $\endgroup$ – Paramanand Singh Nov 9 at 2:54
  • $\begingroup$ @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. $\endgroup$ – user Nov 9 at 3:02
  • 1
    $\begingroup$ @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. $\endgroup$ – Paramanand Singh Nov 9 at 3:04
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ @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. $\endgroup$ – N. S. Nov 8 at 23:08
  • $\begingroup$ 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. $\endgroup$ – N. S. Nov 8 at 23:10
  • 1
    $\begingroup$ @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. $\endgroup$ – Paramanand Singh Nov 9 at 2:23
  • 1
    $\begingroup$ @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. $\endgroup$ – Paramanand Singh Nov 9 at 2:24
  • 1
    $\begingroup$ @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. $\endgroup$ – Paramanand Singh Nov 9 at 2:34
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ is my proof that it's not in an indeterminate form correct? $\endgroup$ – user718615 Nov 9 at 0:03
  • $\begingroup$ @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. $\endgroup$ – user Nov 9 at 0:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy