How to show that $\lim\limits_{x \to 0} \frac{3x-\sin 3x}{x^3}=9/2$?

Question

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

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 $1$.

Answer 1

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,

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

Answer 2

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

Answer 3

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

Answer 4

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

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

Answer 5

Hint: Apply the Hospital rule 3 times you obtain $\lim_{x\rightarrow 0}{{27\cos(3x)}\over 6}={9\over 2}$.

Answer 6

Use L'Hospital's rule:

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

Answer 7

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 :

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