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Find the limit $\displaystyle \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor expansion.

My Try

$\displaystyle =\lim _{x \to 0} \frac {\cos x - \frac{\sin x}{x}} {x \sin x}$

$=\frac {\displaystyle\lim _{x \to 0}\cos x - \lim _{x \to 0}\frac{\sin x}{x}} {\displaystyle\lim _{x \to 0}x \sin x}$

$=\frac{1-1}{0}$

But still I end up with $\frac00$

Any hint for me to proceed would be highly appreciated.

P.S: I did some background check on this question on mathstack and found they have solved this with l'Hopital's rule and the answer seems to be $\frac{-1}{3}$.

What is $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$?

V.G
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emil
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5 Answers5

5

Famously $\lim_{x\to0}\frac{\sin x}{x}=1$ can be proved without such techniques, and implies $\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12$. With a little more effort (e.g. by approximating a circular arc as a parabola), you can also show $\lim_{x\to0}\frac{x-\sin x}{x^3}=\tfrac16$. So $\lim_{x\to0}\frac{x-\sin x}{x^2\sin x}=\tfrac16$ and$$\lim_{x\to0}\left(\frac{\cos x-1}{x\sin x}+\frac{x-\sin x}{x^2\sin x}\right)=-\frac12+\frac16=-\frac13.$$

J.G.
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4

We have

$$\frac {x \cos x - \sin x} {x^2 \sin x}=\frac{\cos x}{\frac{\sin x}x}\cdot\frac{x-\tan x}{x^3}\to1\cdot \left(-\frac13\right)=-\frac13$$

using

user
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    (+1) Rob's answers are always great. – Mark Viola Jul 28 '20 at 18:59
  • @MarkViola I absolutely agree with you, and your also are inspiring. – user Jul 28 '20 at 19:03
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    Oh, wow. Thank you. That comment made my day ... ;- ) – Mark Viola Jul 28 '20 at 19:14
  • @MarkViola There is nothing great about 100k users posting answers to questions a competent freshman can answer. – Jyrki Lahtonen Jul 30 '20 at 09:01
  • @JyrkiLahtonen as a former moderator, I expect you should know that "MSE is a question and answer site for people studying math at any level", refer to TOUR. I have a freshman level, you can't avoid my participation for that. Moreover you should know that we should "Focus on the content, not the person", refer to COC. If you don't like an answer you can freely downvote it but I invite warmly you to avoid these kind of comments otherwise these will be flagged to moderator's attention as infrigiments of our rules. – user Jul 30 '20 at 09:23
  • I'm not targeting any particular user with my downvotes, so I don't see any infringements of rules. I find the practice so lame (as another user phrased it), and it is my right to fight this cancer of the site. – Jyrki Lahtonen Jul 30 '20 at 09:31
  • Hi Jyrki. What on earth are you talking about? I wrote "Rob's answers are always great." I meant that " @robjohn " posts great answers in general, and his posts that "user" referenced in the answer herein are great specifically. Are you implying that robjohn shouldn't have posted such answers simply because of his reputation status on MSE? – Mark Viola Jul 30 '20 at 13:58
  • @MarkViola I don't see Rob answering here. And if you want to be sure that I see your comment, please add the at-sign. Mind you, Robjohn also occasionally answers questions that IMO should be left for new users, or remain unanswered (when the question is the umpteenth reincarnation of the same technique). – Jyrki Lahtonen Aug 02 '20 at 06:00
2

The limit in question is equal to the second derivative of the sinc function evaluated at $0$. That is,

$$\begin{align} \lim_{x\to 0}\frac{x\cos(x)-\sin(x)}{x^2\sin(x)}&=\lim_{x\to 0}\frac{x\cos(x)-\sin(x)}{x^3}\frac{x}{\sin(x)}\\\\ &=2\lim_{h\to 0}\frac{\text{sinc}(h)-1}{h^2}\\\\ &=2\lim_{h\to 0}\frac{\sin(h)-h}{h^3}\\\\ \end{align}$$

In This Answer, I showed, without use of calculus, that the sine function satisfies the inequality

$$\sin(h)\ge h-\frac16 h^3\tag1$$

In a parallel development, one can show, without calculus, that $\sin(h)\le h-\frac16h^3+\frac1{120}h^5$. (Alternatively, integrate $(1)$ twice and use $\cos(0)=1$ and $\sin(0)=0$.)

Hence, applying the squeeze theorem, we find that

$$\lim_{x\to 0}\frac{x\cos(x)-\sin(x)}{x^2\sin(x)}=-\frac13$$

Mark Viola
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  • Thank you for your answer! But I really didn't know what sinc function is, so in fact I didn't understand your proof at my level. But I looked it up on web and came to know that its the same function as $\frac{\sin x}{x}$. This gave me some inspiration actually. I'm willing to learn new areas in mathematics. This will help me in future indeed! – emil Jul 31 '20 at 17:58
  • You're welcome. My pleasure. And glad this was useful. – Mark Viola Jul 31 '20 at 18:19
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While searching more easy way, let me suggest one possible way to solve main difficult part of suggested limit. I change denumerator to $x^3$, for simplicity, as it's equivalent $x^2\sin x$

Suppose we know existence of limit. Then $$L=\lim_{x\to0}\frac{x-\sin x}{x^3} = \lim_{x\to0}\frac{x-3\sin \frac{x}{3}+4 \sin^3 \frac{x}{3}}{x^3}=\\ =\lim_{x\to0}\left(3\frac{\frac{x}{3} - \sin \frac{x}{3}}{x^3} + \frac{4 \sin^3 \frac{x}{3}}{x^3}\right) =\lim_{x\to0}\left(\frac{\frac{x}{3} - \sin \frac{x}{3}}{9\left(\frac{x}{3}\right)^3} + \frac{4 \sin^3 \frac{x}{3}}{x^3}\right)=\frac{L}{9}+\frac{4}{27}$$ From obtained equation $L=\frac{1}{6}$

zkutch
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  • But this is not the limit in question. – Mark Viola Jul 28 '20 at 19:22
  • It is, as I wrote, main part of it: you obtain this hard one after adding-subtracting $x$ in numerator. Look at J.G.'s answer second summand. – zkutch Jul 28 '20 at 19:24
  • The numerator is $x\cos(x)-\sin(x)$, not $x-\sin(x)$. – Mark Viola Jul 28 '20 at 19:26
  • @Mark Viola. I wrote adding-subtracting $x$ in numerator, which is $x\cos(x)-\sin(x) = x\cos(x) -x + x - \sin(x) = x[\cos(x)-1 ] + [x - \sin(x)]$. Now hard part is second, which I wrote above. – zkutch Jul 28 '20 at 19:30
  • Perhaps you should include all of that in your post. In addition, you have assumed that the limit exists, but have not proven that it exists. – Mark Viola Jul 28 '20 at 19:32
  • Perhaps I'll make it. You see, I especially wrote "Suppose we know existence of limit", because I think about ways without derivative. Using derivative seems little unsportsmanlike with tag no l'Hopital, even for monotony. Meanwhile, what is your opinion - we can assume, that initial limit exists? – zkutch Jul 28 '20 at 19:39
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Denote $L$ the existing limit. Then, express it as

\begin{align} L=\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} &= \lim _{x \to 0}\frac {x (2\cos^2\frac x2 -1) -2 \sin \frac x2 \cos\frac x2} {2x^2 \sin \frac x2 \cos\frac x2}\\ &= \lim _{x \to 0} \frac {x (\cos^2\frac x2-1) +2 \cos\frac x2(\frac x2\cos \frac x2- \sin\frac x2)} {2x^2 \sin \frac x2 \cos\frac x2}\\ &= - \lim _{x \to 0}\frac{\sin\frac x2}{\frac x2} \frac1{4\cos\frac x2} + \lim _{x \to 0} \frac{\frac x2\cos \frac x2- \sin\frac x2} {4(\frac x2)^2 \sin \frac x2} \\ &= -\frac14+\frac14L \end{align}

Thus, $L= -\frac13$.

Quanto
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