Solve $\lim _{x\to 0}\left(\frac{1-\cos \left(x\right)\sqrt{\cos \left(2x\right)}\cdot \sqrt[3]{\cos \left(3x\right)}}{x^2}\right)$

Question

$$\lim _{x\to 0}\left(\frac{1-\cos \left(x\right)\sqrt{\cos \left(2x\right)}\cdot \sqrt[3]{\cos \left(3x\right)}}{x^2}\right)$$

Answer 1

HINT:

$$\lim_{x\to0}\dfrac{1-(\cos^6x\cos^32x\cos^23x)^{1/6}}{x^2}$$

$$=\lim_{x\to0}\dfrac{1-\cos^6x\cos^32x\cos^23x}{x^2}\cdot\lim_{x\to0}\dfrac1{1+\sum_{r=1}^5(\cos^6x\cos^32x\cos^23x)^{r/6}}$$

The second limit converges to $\dfrac1{1+\sum_{r=1}^51^{r/6}}=\dfrac16$

For the first, $\cos^6x=(1-\sin^2x)^3=1-3\sin^2x+\cdots$

$\cos^32x=(1-2\sin^2x)^3=1-6\sin^2x+\cdots$

and $\cos^23x=\dfrac{1+\cos6x}2=\dfrac{1+4\cos^32x-3\cos2x}2=\dfrac{1+4\{(1-2\sin^2x)\}^3-3(1-2\sin^2x)}2=1-9\sin^2x+\cdots$

as $\cos2x=1-2\sin^2x$

Answer 2

Hint:

• $\cos(x)=1-\frac{x^2}{2}+o(x^2)$ as $x\to0$,
• $(1+t)^a=1+at+o(t)$ as $t\to0$

Answer 3

Easy to know: $$(\cos kx)^{1/k}=1-k·x^2/2+o(x^2).$$ Obviously $$1-\cos x(\cos（2x）^1/2)(\cos（3x）^1/3)=1-(1-x^2/2)(1-x^2)(1-3x^2/2)+o(x^2).$$ So the limit $=1/2+1+3/2=3$.

Answer 4

Near $x=0$, we have $$\cos(x)=1-\frac{x^2}{2}+O\left(x^4\right),$$ \begin{align}\sqrt{\cos(2x)} &= \sqrt{1-\frac{4x^2}{2}+O\left(x^4\right)} \\ &= 1-\frac{1}{2}\frac{4x^2}{2}+O\left(x^4\right), \end{align} \begin{align}\left(\cos(3x)\right)^{1/3} &= \left(1-\frac{9x^2}{2}+O\left(x^4\right)\right)^{1/3} \\ &= 1-\frac{1}{3}\frac{9x^2}{2}+O\left(x^4\right), \end{align} and then \begin{align}\lim_{x\to 0}\frac{1-\cos(x)\sqrt{\cos(2x)}\left(\cos(3x)\right)^{1/3}}{x^2} &= \lim_{x\to 0}\frac{1-\left(1-\frac{x^2}{2}+O\left(x^4\right)\right)\left(1-x^2+O\left(x^4\right)\right)\left(1-\frac{3x^2}{2}+O\left(x^4\right)\right)}{x^2} \\ &= \lim_{x\to 0}\frac{1- \left(-\frac{x^2}{2}-x^2-\frac{3x^2}{2}+O\left(x^4\right)\right)}{x^2} \\ &= \lim_{x\to 0}\frac{3x^2+O\left(x^4\right)}{x^2} \\ &= 3+\lim_{x\to 0}O\left(x^2\right) \\ &= 3. \end{align}