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$\lim\limits_{x \to 0} \frac{3x-\sin 3x}{x^3}$

I need to prove that this limit equals to $\frac{9}{2}$. Can someone give me a step by step solution?

EDIT: I am sorry. The $x$ goes to $0$, not $1$.

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If you allow Taylor expansions, recall that

$$\sin(x)=x-\frac16x^3+\mathcal O(x^5)$$

Thus,

$$\sin(3x)=3x-\color{red}{\frac92}x^3+\mathcal O(x^5)$$

Thus,

$$\begin{align}\frac{3x-\sin(3x)}{x^3}&=\frac{\frac92x^3+\mathcal O(x^5)}{x^3}\\&=\frac92+\mathcal O(x^2)\\&\to\frac92\end{align}$$

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  • $\begingroup$ Very nice solution, thanks a lot. $\endgroup$ – Thomas Feb 28 '17 at 23:50
  • $\begingroup$ Taylor expansions make everything easy especially this limit as this limit has a question dedicated to it for proving without Taylor or L'hoptial. $\endgroup$ – A---B Mar 1 '17 at 0:02
  • $\begingroup$ @A---B Indeed, but it's not to hard to derive the Taylor expansion either. Once you know the basic trig limits, deriving derivative of derivatives is not hard at all. $\endgroup$ – Simply Beautiful Art Mar 1 '17 at 0:09
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Applying L'Hopital's rule three times $$\lim_{x \to 0}\frac{3x-\sin(3x)}{x^3}=\lim_{x \to 0}\frac{3-3\cos(3x)}{3x^2}=\lim_{x \to 0}\frac{9\sin(3x)}{6x}=\lim_{x \to 0}\frac{27\cos(3x)}{6}=\frac{9}{2}.$$

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Using l'Hôpital's rule;$$\lim_{x\to0}\frac{3x-\sin(3x)}{x^3}=\lim_{x\to0}\frac{3-3\cos(3x)}{3x^2}=\lim_{x\to0}\frac{9\sin(3x)}{6x}=\lim_{x\to0}\frac{27\cos(3x)}{6}=\frac{9}{2}$$

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  • $\begingroup$ Yeah i tried to use l'Hôpital's rule but I did something wrong. Thanks for helping me out $\endgroup$ – Thomas Feb 28 '17 at 23:51
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Hint: Apply the Hospital rule 3 times you obtain $\lim_{x\rightarrow 0}{{27\cos(3x)}\over 6}={9\over 2}$.

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    $\begingroup$ No offense, but I think this is rather low quality. $\endgroup$ – Simply Beautiful Art Feb 28 '17 at 23:38
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    $\begingroup$ Why so? Not agreeing or disagreeing, but could you elaborate just a bit? $\endgroup$ – Harnoor Lal Feb 28 '17 at 23:46
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By elementary means:

From

$$\sin 3x=3\sin x-4\sin^3x$$

we draw

$$L=\lim_{x\to0}\frac{3x-3\sin x+4\sin^3x}{x^3}=\lim_{x\to0}\frac{3x-3\sin x}{x^3}+4.$$

But

$$\lim_{x\to0}\frac{x-\sin x}{x^3}=\lim_{3x\to0}\frac{3x-\sin3x}{27x^3}$$ so that

$$L=\frac L9+4.$$

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    $\begingroup$ This doesn't prove that $L$ exists, does it? $\endgroup$ – user1551 Mar 1 '17 at 12:43
  • $\begingroup$ @user1551: that's right, it only proves that if it exists, it has value $9/2$. $\endgroup$ – Yves Daoust Mar 1 '17 at 13:10
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Use L'Hospital's rule:

$$\lim _{x\to0} \frac{3x-\sin3x}{x^3} = \lim_{x\to0} \frac{3-3\cos3x}{3x^2} = \lim _{x\to0} \frac{9\sin3x}{6x} = \lim_{x\to0} \frac{27\cos3x}{6} = \frac{27}{6} = \frac{9}{2}$$

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    $\begingroup$ There is a symbol \to in $\LaTeX$ which renders as a right arrow: $\to$ and serves for limits instead of a 'minus'+'greater than' pair. $\endgroup$ – CiaPan Mar 1 '17 at 11:24
  • $\begingroup$ I am rather new to LATEX so I am not quite acclimatized with it.Please bear with me. $\endgroup$ – Saradamani Mar 2 '17 at 4:01
  • $\begingroup$ Don't worry, everyone was once a beginner :) $\endgroup$ – CiaPan Mar 2 '17 at 21:13
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first :

theorem :

let $f(0)=0 , f'(0)$ have existed then :

$$\lim_{x\to 0} \frac{f(x)-\sin(f(x))}{x^3}= \frac{1}{6}(f'(0))^3$$

now :

$$\lim_{x\to 0} \frac{3x-\sin(3x)}{x^3}= \frac{1}{6}(3)^3=\frac{9}{2}$$

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