$\lim_{x\to 0} \frac{x\sin(3x)}{1-\cos(6x)}$

Question

$$\lim_{x\to 0} \frac{x\sin(3x)}{1-\cos(6x)}$$

I tried the following but it doesn't seem to work...

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

But the result of this limit is $\frac{1}{6}$.

Am I missing something or did I make any glaring mistakes?

Answer 1

We have that

$$\frac{x\sin(3x)}{1-\cos(6x)}=\frac{\color{red}{(6x)^2}}{1-\cos(6x)}\cdot\frac{\sin(3x)}{3x}\cdot \frac1{12}\to 2\cdot \ 1\cdot \frac1{12}=\frac16$$

Answer 2

$$\dfrac{x\sin 3x}{1-\cos 6x}=\dfrac{x\sin 3x}{2\sin^2 3x}=\dfrac{x}{2\sin 3x}.$$ Therefore, $$\lim_{x\to 0}\dfrac{x\sin 3x}{1-\cos 6x}=\lim_{x\to 0}\dfrac{1}{6\times \dfrac{\sin 3x}{3x}}=\dfrac{1}{6}.$$

Answer 3

Your mistake is that the second limit is not $0$. It's an indeterminate form $0\times \infty$. Since $\lim_{x\to 0}\frac{1-\cos(6x)}{6x} = 0$, $\lim_{x\to 0}\frac{6x}{1-\cos(6x)}$ doesn't exist (diverges to $\pm \infty$) and you also have $\lim_{x\to 0}\frac{x}{2} = 0$. Use one of the methods in the other answers for the correct solution.

Answer 4

More generally:

All you need is $\dfrac{\sin(x)}{x} \to 1$ and $1-\cos(x) =2\sin^2(x/2) $.

$\begin{array}\\ \dfrac{x\sin(ax)}{1-\cos(bx)} &=ax^2\dfrac{\sin(ax)}{ax}\dfrac{1}{2\sin^2(bx/2)}\\ &=\dfrac{ax^2}{2(b/2)^2}\dfrac{\sin(ax)}{ax}\dfrac{(b/2)^2}{\sin^2(bx/2)}\\ &=\dfrac{a}{b^2/2}\dfrac{\sin(ax)}{ax}\dfrac{(bx/2)^2}{\sin^2(bx/2)}\\ &\to \dfrac{2a}{b^2}\\ \end{array} $

For $a=3, b=6$ this is $\dfrac{2\cdot 3}{6^2} =\dfrac16 $.

Answer 5

We can also make the substitution $x \mapsto \frac{u}{3}$:

$$\frac{1}{3} \frac{u \sin u}{1 - \cos 2u} = \frac{1}{3} \frac{u \sin u}{1 - \cos^2 u + \sin^2 u} = \frac{1}{3} \frac{u \sin u}{2 \sin^2 u} = \frac{1}{6}\frac{u}{\sin u}.$$

and since $\frac{u}{\sin u} \to 1$ as $u \to 0$:

$$\lim_{u \to 0} \frac{1}{3} \frac{u \sin u}{1- \cos 2u} = \frac{1}{6} \times 1 = \frac{1}{6}.$$