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What's the value of this limit? I keep getting half, but the answer's 1/3. I think there might be a problem with writing ${x\cot x \over x^2 }$ as $ {\cos x \over x^2}$. Is there ? I get (1-cosx)/x^2 on using this substitution, and if I use L'hôpital's Rule on this, I get the answer as sinx/2x = 1/2.

Edit: I can only use results like $\sin x=x$ and L'Hospital's rule. No series expansion.

Edit 2: Here's my attempt:

1/x^2 - cosx/(x*sinx) = 1/x^2 - (cosx/x^2)(x/sinx) = 1/x^2 - (cosx/x^2)(1) {Since sinx/x=1} = (1-cosx)/x^2 Being a 0/0 case, L'hôpital's Rule: sinx/2x = (1/2)(sinx/x) = 1/2, since sinx/x=1

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  • $\begingroup$ $\sin x$ is not really equal to $x$, it is $x+o(x)$, so you can write $\sin x\sim_0x$, which will work well and relies only, in this case, on the high school limit: $\frac{\sin x}x\to 1$ as $x\to 0$. $\endgroup$
    – Bernard
    Dec 26, 2019 at 10:18

6 Answers 6

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Rewrite the limit as $$ \lim_{x\to0}\frac{\sin x-x\cos x}{x^2\sin x}=\lim_{x\to0}\frac{\sin x-x\cos x}{x^3}\frac{x}{\sin x} $$ The second fraction has limit $1$, so you just need to compute $$ \lim_{x\to0}\frac{\sin x-x\cos x}{x^3}=\lim_{x\to0}\frac{\cos x-\cos x+x\sin x}{3x^2}=\lim_{x\to0}\frac{1}{3}\frac{\sin x}{x} $$

Your attempt is faulty, because $$ \lim_{x\to0}\frac{1}{x^2}=\infty,\qquad \lim_{x\to0}\frac{\cot x}{x}=\infty $$ so you have a form $\infty-\infty$ and you can't use that $\sin x=x$ (which is false, by the way, avoid thinking like that).

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  • $\begingroup$ But you've implicitly used sinx = x here, isn't that illegal because you have a third power term in the denominator? Won't you have to take a few more terms from the expansion? $\endgroup$
    – Arnav Das
    Dec 26, 2019 at 17:02
  • $\begingroup$ @ArnavDas I use the theorem on product of limits. Since $\lim_{x\to0}\frac{\sin x-x\cos x}{x^3}=1/3$ (with l'Hôpital) and $\lim_{x\to0}x/\sin x=1$, the limit of the product is $1/3$. $\endgroup$
    – egreg
    Dec 26, 2019 at 17:05
  • $\begingroup$ Can the limit x tends to 0 be split across a + or - sign? $\endgroup$
    – Arnav Das
    Dec 26, 2019 at 17:14
  • $\begingroup$ @ArnavDas It depends: the important thing is that each part has a finite limit (or both $+\infty$, both $-\infty$ or one finite and the other infinite). Surely you cannot split the given limit as $\lim_{x\to0}1/x^2-\lim_{x\to0}(\cot x)/x$. I added a comment on your attempt. $\endgroup$
    – egreg
    Dec 26, 2019 at 17:17
  • $\begingroup$ Can you explain why I can't split it? Didn't you give me a mutually exhaustive set of cases there, given one of them is infinite?Here, 1/x^2 is infinite, so doesn't that mean I can split it? $\endgroup$
    – Arnav Das
    Dec 26, 2019 at 17:21
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Hint

Use the series expansion $$\cot(x)=\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}+O\left(x^5\right)$$ what could also be obtained from $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$ followed by long division.

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  • $\begingroup$ I need a solution without expansion, but only using standard results like sinx=x=tanx. Is there something wrong with taking xcotx/x^2 as cosx/x^2 $\endgroup$
    – Arnav Das
    Dec 26, 2019 at 10:03
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Since $\lim_{x\to0}\frac{\tan x}{x}=1$ and $\lim_{x\to0}\frac{\tan x-x}{x^3}=\frac13$, $$\lim_{x\to0}\frac{1-x\cot x}{x^2}=\lim_{x\to0}\frac{\tan x-x}{x^2\tan x}=\frac{\frac13}{1}=\frac13.$$

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  • $\begingroup$ Where does the second result come from? Is this standard? $\endgroup$
    – Arnav Das
    Dec 26, 2019 at 10:05
  • $\begingroup$ @ArnavDad One of several worth knowing. $\endgroup$
    – J.G.
    Dec 26, 2019 at 10:07
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Hint:

$$\dfrac{1-x\cot x}{x^2}=\dfrac{\tan x-x}{x^3}\cdot\dfrac x{\tan x}$$

Use Are all limits solvable without L'Hôpital Rule or Series Expansion

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  • $\begingroup$ Okay, so here if I take the term on the right (tanx/x) as 1 and then take tanx=x, won't I get zero? What's wrong with this? Am I doing something against the rules? $\endgroup$
    – Arnav Das
    Dec 26, 2019 at 10:04
  • $\begingroup$ @ArnavDas, It will be $$\dfrac00$$ right? $\endgroup$ Dec 26, 2019 at 10:18
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Rewrite this fraction first: $$ {1-x\cot x\over x^2}=\frac{\tan x -x}{x^2\tan x}$$ and use Taylor expansion at order $3$ for the tangent: $$\tan x=x+\frac{x^3}3+o(x^3),\quad\text{so }\quad \tan x- x =\frac{x^3}3+o(x^3)\sim_0 \frac{x^3}3.$$ Therefore, as $\tan x\sim_0 x$, $$\frac{\tan x -x}{x^2\tan x}\sim_0 \frac{\frac{x^3}3}{x^3}=\frac13.$$

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  • $\begingroup$ Can you check my edit? $\endgroup$
    – Arnav Das
    Dec 26, 2019 at 10:03
  • $\begingroup$ @ArnavDas: It's fine. What was my typo (or were…)? $\endgroup$
    – Bernard
    Dec 26, 2019 at 10:06
  • $\begingroup$ I meant my edit on the question. $\endgroup$
    – Arnav Das
    Dec 26, 2019 at 10:06
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    $\begingroup$ Oh! I see. Well, you implicitly used that near $0$, $\sin x\sim x$ (in the sense of asymptotic analysis) to simplify the numerator, but you should explain in a few words what you're doing, in my opinion. $\endgroup$
    – Bernard
    Dec 26, 2019 at 10:11
  • $\begingroup$ How do I know which order do I have to stop at? You've taken terms till power 3, is that because the denominator was x^2? $\endgroup$
    – Arnav Das
    Dec 26, 2019 at 17:12
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By L'Hospital and using $\dfrac{\sin x}x\to1$,

$$\lim_{x\to0}\frac{\sin x-x\cos x}{x^3}=\lim_{x\to0}\frac{x\sin x}{3x^2}=\frac13.$$

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  • $\begingroup$ In what does this differ from my answer? $\endgroup$
    – egreg
    Dec 26, 2019 at 17:36

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