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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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    Is it supposed to be $x\to\color{red}1$? – 2017-02-28
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    Are you sure that you wrote the correct thing? That limit isn't 9/2. – 2017-02-28
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    http://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhôpital-rule-or-series-expansion – 2017-03-01

7 Answers 7

10

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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    Very nice solution, thanks a lot. – 2017-02-28
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    Taylor expansions make everything easy especially this limit as this limit has a question dedicated to it for proving without Taylor or L'hoptial. – 2017-03-01
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    @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. – 2017-03-01
6

Applying L'Hopital's rule three times $$\lim_{x \to 0}\frac{3x-\sin(3x)}{x^3}=\lim_{x \to 0}\frac{3-3\cos(3x)}{3x^2}=\lim_{x \to 0}\frac{9\sin(3x)}{6x}=\lim_{x \to 0}\frac{27\cos(3x)}{6}=\frac{9}{2}.$$

2

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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    Yeah i tried to use l'Hôpital's rule but I did something wrong. Thanks for helping me out – 2017-02-28
1

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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    No offense, but I think this is rather low quality. – 2017-02-28
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    Why so? Not agreeing or disagreeing, but could you elaborate just a bit? – 2017-02-28
1

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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    This doesn't prove that $L$ exists, does it? – 2017-03-01
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    @user1551: that's right, it only proves that if it exists, it has value $9/2$. – 2017-03-01
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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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    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. – 2017-03-01
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    I am rather new to LATEX so I am not quite acclimatized with it.Please bear with me. – 2017-03-02
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    Don't worry, everyone was once a beginner :) – 2017-03-02
-1

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 :

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