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The problem goes : Solve the following limit without using l'hopital's rule : $$\lim\limits_{x\to 0} \frac{x-\sin(x)}{x^3}.$$ I've tried multiplying with conjugate "$x+\sin(x)$", I've tried extracting $x$ from the numerator, and I always end up in a dead end. Pleas help.

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marked as duplicate by Martin R, Simon S, Rob Arthan, muaddib, Mark Viola Aug 13 '15 at 21:26

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    $\begingroup$ To preempt answers that try to sneak in the derivative other ways: can you use power series approximations? I.e., $\sin x = x - x^3/3! + O(x^5)$? $\endgroup$ – Simon S Aug 13 '15 at 20:32
  • $\begingroup$ It is often included as a theorem or fact that $\lim\limits_{x\to 0}\frac{\sin(x)}{x} = 1$, frequently at the same time or just before L'Hopital's rule is introduced. Do you have access to that result? In Calculus 7th ed. by Larson, Hostetler, and Edwards, this result appears as theorem 1.9 on page 63. L'Hopital's rule is not introduced until page 531. $\endgroup$ – JMoravitz Aug 13 '15 at 20:35
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    $\begingroup$ This seems like a farce of a problem. If Taylor approximations are available, this automatically gives L'Hopital's rule as a consequence with very little extra work. If Taylor AND L'Hoptial are not allowed, how can one proceed? $\endgroup$ – user2566092 Aug 13 '15 at 20:40
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Just expand $\sin(x)$ in a power series and you are done:

$$ \lim_{x\rightarrow 0} \frac{x-\sin(x)}{x^3} = \lim_{x\rightarrow 0} \frac{x-(x-x^3/3!+\mathcal{O}(x^5))}{x^3} =\lim_{x\rightarrow 0} (1/3!+\mathcal{O}(x^2)) =1/6 $$

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    $\begingroup$ Taylor is equivalent to L'Hôpital. $\endgroup$ – Vincenzo Oliva Aug 13 '15 at 20:55
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I offer an alternative route based on a trigonometric identity.

You will need to prove that the limit $L$ exists.

Put $x=3u$ in $$L = \lim_{x\to0}{\frac{x-\sin x}{x^3}}$$

to get

$$L = \lim_{u\to 0}{\frac{3u-\sin(3u)}{(3u)^3}}$$

and since $$\sin(3u)=-4\sin^3 u + 3\sin u$$

we get $$\begin{align} L = \lim_{u\to 0}{\frac{3u-(-4\sin^3 u + 3 \sin u)}{27u^3}} &= \lim_{u\to 0}{\left\{\frac{4}{27}\cdot\left(\frac{\sin u}{u}\right)^3 + \frac{u-\sin u}{9u^3}\right\}} \\\\ &= \frac{4}{27}\left(\lim_{u\to 0}{\frac{\sin u}{u}} \right)^3 + \frac{1}{9}\left(\lim_{u\to 0}{\frac{u-\sin u}{u}}\right) \\\\ &= \frac{4}{27}\cdot1 + \frac{1}{9}L\qquad(\textit{if given }\lim_{u\to 0}{\frac{\sin u}{u}} = 1) \end{align}$$

So $$\frac{8}{9}L=\frac{4}{27} \implies L = \frac{9}{8}\cdot\frac{4}{27} = \frac{1}{6}$$

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We have using integration by parts

$$\sin(x)=\int_0^x\cos(t)dt=x-\int_0^x(x-t)\sin(t)dt$$ and by integrating twice by parts we get $$\sin x=x-\frac{x^3}{6}+\int_0^x\frac{(x-t)^3}{6}\sin(t)dt$$ Finally since $$\left|\int_0^x\frac{(x-t)^3}{6}\sin(t)dt\right|\le \int_0^x\frac{(x-t)^3}{6}dt=\frac{x^4}{24}$$ we get easily the desired limit $\frac16$.

Remark : This method is in fact the proof of the Taylor series of the function $\sin$ with integral remainder.

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    $\begingroup$ Are you disguising Taylor? $\endgroup$ – Vincenzo Oliva Aug 13 '15 at 20:56
  • $\begingroup$ Yes I'm, Isn't this allowed? $\endgroup$ – user260717 Aug 13 '15 at 21:20
  • $\begingroup$ Well, Taylor is equivalent to L'Hôpital. $\endgroup$ – Vincenzo Oliva Aug 13 '15 at 22:20

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