Evaluate $$\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right).$$

Attempt \begin{align*} &\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right)\\ = &\lim_{x \to 0} \left(\frac{1}{x}-\cot{x}\right)\left(\frac{1}{x}+\cot{x}\right)\\ = &\lim_{x \to 0} \left(\frac{\sin{x}+x\cos{x}}{x\sin{x}}\right)\left(\frac{\sin{x}-x\cos{x}}{x\sin{x}}\right)\\ = &\lim_{x \to 0} \left(\frac{\sin{x}+x\cos{x}}{x\sin{x}}\right) \times \lim_{x \to 0}\left(\frac{\sin{x}-x\cos{x}}{x\sin{x}}\right). \end{align*}

Both the terms are in $\frac00$ form. So applying L'Hopital on both the limits we have,

$$= \lim_{x \to 0} \left(\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}\right) \times \lim_{x \to 0}\left(\frac{x\sin{x}}{x\cos{x}+\sin{x}}\right).$$

The second term is in $\frac00$ form. So applying L'Hopital on the second limit we have, \begin{align*} = &\lim_{x \to 0} \left(\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}\right) \times \lim_{x \to 0}\left(\frac{x\cos{x}+\sin{x}}{2\cos{x}-x\sin{x}}\right)\\ =& \lim_{x \to 0} \left(\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}\right) \times \left(\frac{x\cos{x}+\sin{x}}{2\cos{x}-x\sin{x}}\right)\\ =& 1 \end{align*}

The correct answer is $\dfrac23$ which can be found using series expansion. But I think I'm making a conceptual mistake in one of the above steps. Could you please point out to the specific step where I've committed a mistake in above solution?

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    $\begingroup$ The = on the last 2nd line. Generally speaking, you could do that if both the limit exists. In your case, that is not fully justified. $\endgroup$ – xbh Oct 24 '18 at 8:21
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    $\begingroup$ @xbh: Please post your comment as an answer, because not a single one of the four existing answers have identified the error, whereas you have. So none of them have addressed the question (point to the specific step with a mistake). $\endgroup$ – user21820 Oct 24 '18 at 14:03
  • $\begingroup$ +1 for the nice question. For beginners it is really painful to figure out such hidden errors. $\endgroup$ – Paramanand Singh Oct 24 '18 at 23:22
  • $\begingroup$ The 4th line is also wrong. The first limit converges to $\infty$ and the second to $0$. So you can't say the limit of the product is the product of the limits. $\endgroup$ – Kitegi Oct 28 '18 at 10:43

As @user21820 encouraged, I am going to explain the errors in this post.

The problem is the $=$ at the last 2nd line. Before this everything could be accepted, since they are the process to find the limits. But at this step you generally claim that $$ \lim f(x) \lim g(x) = \lim f(x)g(x), $$ where the limits progression are omitted for brevity. By the arithmetic operation of limits, we know that when $\lim f(x), \lim g(x)$ exists, then the equation above holds. In your case, the existence of $\lim f, \lim g$ are not examined at all. Then it is not valid to combine two limits and make it to one limit, since there is a counterexample $$ 1=\lim_{x \to 0} \frac xx , \lim _{x\to 0}x =0, \lim_{x\to 0}\frac 1x \text{ does not exist}, $$ where you obviously cannot write $1 = 0 \times \lim_{x \to 0}(1/x)$. As other answers showed, not all the two parts $$ \lim_{x\to 0} \frac {2\cos x -x \sin x}{x \cos x + \sin x}, \lim_{x\to 0} \frac {x \cos x + \sin x} {2\cos x -x \sin x} $$ exist, so the certain line makes no sense, and leads you to the wrong result.

  • $\begingroup$ Thank you. This clarification is exactly what I needed. $\endgroup$ – yathish Oct 24 '18 at 15:05
  • $\begingroup$ Yup, and it's good to see that you also distinguish the "process to find the limits" and the "justification of the limits", as I did in my comment. $\endgroup$ – user21820 Oct 24 '18 at 15:06
  • $\begingroup$ @yathish: I probably should add that if you want to be rigorous, you can append "if the limits involved (on this line) exist" to every line. Then everything is correct up until the second-last equality. =) $\endgroup$ – user21820 Oct 24 '18 at 15:09
  • $\begingroup$ @user21820 Right. Thanks for your time and prodding xbh to post an answer. $\endgroup$ – yathish Oct 24 '18 at 15:11
  • $\begingroup$ The problem is not only in combining the limits at the end but also while splitting in the beginning. For product of limits these steps are valid if one of the factors has a non-zero limit. But sadly that's not the case here. The approach in question never actually evaluates the limit of any factors and manages to combine them to produce $1$ finally and hence the error remains hidden. $\endgroup$ – Paramanand Singh Oct 24 '18 at 23:05

\begin{align}\lim_{x\to0}\frac1x+\cot x&=\lim_{x\to0}\frac{\sin x+x\cos x}{x\sin x}\\&=\lim_{x\to0}\frac{2\cos x-x\sin x}{\sin x+x\cos x}\\&=\infty.\end{align}Besides,$$\lim_{x\to0}\frac1x-\cot x=0.$$

The error you made lies in the use of the equality$$\lim_{x\to a}\bigl(f(x)g(x)\bigr)=\left(\lim_{x\to a}f(x)\right)\left(\lim_{x\to a}g(x)\right)$$in a situation where you cannot apply it (it is the second of your equalities counting from the bottom).

  • $\begingroup$ When solving a problem analytically, how would we know beforehand whether Limit of one of f(x) or g(x) is going to be non-existent? Do we have to check it every time we do the distributive multiplication? $\endgroup$ – yathish Oct 24 '18 at 14:52
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    $\begingroup$ You don't know beforehand. You have to check whether the limits of $f$ and of $g$ exist. If they do and if you are not in a situation in which one of them is $0$ and the other one is $\infty$, then the limit of the product is the product of the limits. $\endgroup$ – José Carlos Santos Oct 24 '18 at 14:56
  • $\begingroup$ Ok. One more clarification, @JoséCarlosSantos. According to this principle, wouldn't my error be in the 4th step itself (counting from top), where I distributed Lim over multiplication? Although we get to know of the incorrectness of the step in hindsight.. $\endgroup$ – yathish Oct 24 '18 at 15:00
  • $\begingroup$ @yathish: To answer your last question, of course. The theorem only asserts that the limit of the product is the product of the limits if the two individual limits exist and their product is well-defined (so that it works for the affinely extended reals too). Obviously, if that condition is not satisfied you cannot use that theorem. $\endgroup$ – user21820 Oct 24 '18 at 15:03
  • $\begingroup$ @yathish: Yes you can say that the splitting step is already wrong, because you won't be able to justify it. However, from a computational perspective it's what you'd want to try anyway. $\endgroup$ – user21820 Oct 24 '18 at 15:04

Both the terms are in 0/0 form. So applying L'Hospital on both the limits we have,

$= \lim_{x \to 0} (\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}) \times \lim_{x \to 0}(\frac{x\sin{x}}{x\cos{x}+\sin{x}})$

Note that the limit $\lim_{x \to 0} (\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}})$ does not exist.

In fact, even before that $\lim_{x \to 0} (\frac{\sin{x}+x\cos{x}}{x\sin{x}})$ does not exist as well.

This is the graph of $\frac{\sin{x}+x\cos{x}}{x\sin{x}}$.

enter image description here

  • $\begingroup$ Thanks. This helped a bit! $\endgroup$ – yathish Oct 24 '18 at 15:07

Similar work: $$\begin{align}\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right)&=\lim_{x \to 0} \left(\frac{1}{x^2}-\frac{1-\sin^2x}{\sin^2x}\right)=\\ &=\lim_{x \to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2x}+1\right)=\\ &=\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^2\sin^2x}\right)+1=\\ &=\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^2\sin^2x}\cdot \frac{\sin^2x}{x^2}\right)+1=\\ &=\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^4}\right)+1. \end{align}$$ According to the algebraic limit theorem, you can express the limit as a product of two existing limits: $$\begin{align}\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^4}\right)+1=&\\ \underbrace{\lim_{x \to 0}\left(\frac{\sin x-x}{x^3}\right)}_{-\frac16}\cdot \underbrace{\lim_{x \to 0}\left(\frac{\sin x+x}{x}\right)}_{=2} +1\stackrel{LR}{=}&\\ \lim_{x \to 0}\left(\frac{\cos x-1}{3x^2}\right)\cdot \lim_{x \to 0}\left(\frac{\cos x+1}{1}\right) +1\stackrel{LR}{=}&\\ \lim_{x \to 0}\left(\frac{-\sin x}{6x}\right)\cdot 2 +1=&\\ -\frac13+1=&\frac23.\end{align}$$ However you can not express: $$-\frac13=\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^4}\right)=\\ \underbrace{\lim_{x \to 0}\left(\frac{\sin x-x}{x^\color{red}0}\right)}_{=0}\cdot \underbrace{\lim_{x \to 0}\left(\frac{\sin x+x}{x^\color{blue}4}\right)}_{=\infty} \ \ \text{OR}\\ \underbrace{\lim_{x \to 0}\left(\frac{\sin x-x}{x^\color{red}1}\right)}_{=0}\cdot \underbrace{\lim_{x \to 0}\left(\frac{\sin x+x}{x^\color{blue}3}\right)}_{=\infty} \ \ \text{OR}\\ \underbrace{\lim_{x \to 0}\left(\frac{\sin x-x}{x^\color{red}2}\right)}_{=0}\cdot \underbrace{\lim_{x \to 0}\left(\frac{\sin x+x}{x^\color{blue}2}\right)}_{=\infty} \ \ \text{OR}\\ \underbrace{\lim_{x \to 0}\left(\frac{\sin x-x}{x^\color{red}4}\right)}_{=\infty}\cdot \underbrace{\lim_{x \to 0}\left(\frac{\sin x+x}{x^\color{blue}0}\right)}_{=0}.$$ because all are the indeterminate form of $0\cdot \infty$.

  • $\begingroup$ Thanks. This was helpful! $\endgroup$ – yathish Oct 24 '18 at 15:06
  • $\begingroup$ Welcome. Good luck. $\endgroup$ – farruhota Oct 24 '18 at 16:56
  • $\begingroup$ @J.G. Thank you for editing the link: it is convenient to see the thereom at once (it was simple: add hashtag and the paragraph heading). Interesting to note we got a pair in our reputations (16.6K). $\endgroup$ – farruhota Oct 24 '18 at 17:26

$$\lim_{x\to0}\left(\dfrac1{x^2}-\dfrac1{\tan^2x}\right)=\lim_{x\to0}\dfrac{\tan x-x}{x^3}\cdot\lim_{x\to0}\dfrac{\tan x+ x}x\cdot\left(\lim_{x\to0}\dfrac x{\tan x}\right)^2$$

Now the limit for the last two are too simple for L'hospital

For $\lim_{x\to0}\dfrac{\tan x-x}{x^3}$ either use L'hospital or Are all limits solvable without L'Hôpital Rule or Series Expansion

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    $\begingroup$ I don't think this fully answers OP's question. $\endgroup$ – 高田航 Oct 24 '18 at 8:22

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