using a trick to solve $\lim_{x \to \ 0}\frac{1-\cos\left(x\right)\sqrt{\cos\left(2x\right)}\sqrt[3]{\cos\left(3x\right)}}{x^{2}}$

Question

This question has been posted once by another user, but the idea used to solve this was not very good and also took much time, I'm asking this again because I've solved the limit by another way but it still is not what I want. here is my solution:

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

$$=\lim_{x \to \ 0}\frac{\sin^{2}\left(x\right)+\cos^{2}\left(x\right)-\cos\left(x\right)\sqrt{\cos\left(2x\right)}\sqrt[3]{\cos\left(3x\right)}}{x^{2}}$$$$=\lim_{x \to \ 0}\left(\frac{\sin\left(x\right)}{x}\right)^{2}+\lim_{x \to \ 0}\frac{\cos\left(x\right)\left(\cos\left(x\right)-1+1-\sqrt{\cos\left(2x\right)}\sqrt[3]{\cos\left(3x\right)}\right)}{x^{2}}$$$$=\lim_{x \to \ 0}\left(\frac{\sin\left(x\right)}{x}\right)^{2}+\lim_{x \to \ 0}\cos\left(x\right)\cdot\frac{\cos\left(x\right)-1+1-\sqrt{\cos\left(2x\right)}\sqrt[3]{\cos\left(3x\right)}}{x^{2}}$$$$=\lim_{x \to \ 0}\left(\frac{\sin\left(x\right)}{x}\right)^{2}+\lim_{x \to \ 0}\cos\left(x\right)\cdot\left(\frac{\cos\left(x\right)-1}{x^{2}}+\frac{1-\sqrt{\cos\left(2x\right)}\sqrt[3]{\cos\left(3x\right)}}{x^{2}}\right)$$$$=\lim_{x \to \ 0}\left(\frac{\sin\left(x\right)}{x}\right)^{2}+\lim_{x \to \ 0}\cos\left(x\right)\cdot\left(\frac{\cos\left(x\right)-1}{x^{2}}+\color{red}{\frac{1-\cos^{3}\left(2x\right)\cos^{2}\left(3x\right)}{x^{2}}}\cdot\frac{1}{1+\sqrt{\cos\left(2x\right)}\sqrt[3]{\cos\left(3x\right)}}\cdot\frac{1}{\left(1+\cos\left(2x\right)\sqrt[3]{\cos^{2}\left(3x\right)+}\cos^{2}\left(2x\right)\sqrt[3]{\cos^{4}\left(3x\right)}\right)}\right)=$$

also we have: $$\color{red} {\frac{1-\cos^{3}\left(2x\right)\cos^{2}\left(3x\right)}{x^{2}}}$$$$=\frac{1}{16x^{2}}(\color{blue}{15}-6\cos(2x)-3\cos(4x)-2\cos(6x)-3\cos(8x)-\cos(12x))$$$$=\frac{1}{16x^{2}}\left(-6\cos(2x)+\color{blue}{6}-3\cos(4x)+\color{blue}{3}-2\cos(6x)+\color{blue}{2}-3\cos(8x)+\color{blue}{3}-\cos(12x)+\color{blue}{1}\right)$$$$=\color {green}{\frac{1}{16}\left(\frac{6\left(-\cos(2x)+1\right)}{x^{2}}+\frac{3\left(-\cos(4x)+1\right)}{x^{2}}+\frac{2\left(-\cos(6x)+1\right)}{x^{2}}+\frac{3\left(-\cos(8x)+1\right)}{x^{2}}+\frac{\left(-\cos(12x)+1\right)}{x^{2}}\right)}$$

using $$\lim_{x \to \ 0}\frac{\left(-\cos(ax)+1\right)}{x^{2}}=\lim_{x \to \ 0}\frac{a^{2}\sin^{2}\left(ax\right)}{a^{2}x^{2}}\cdot\frac{1}{\cos(ax)+1}= a^{2} \lim_{x \to \ 0}\left(\frac{\sin\left(ax\right)}{ax}\right)^{2}\cdot\frac{1}{\cos(ax)+1}=\frac{a^{2}}{2}$$ we have:

$$=\color{green}{\frac{1}{16}\left(6\left(\frac{4}{2}\right)+3\left(\frac{16}{2}\right)+2\left(\frac{36}{2}\right)+3\left(\frac{64}{2}\right)+\left(\frac{144}{2}\right)\right)}=15$$

finally :

$$=\lim_{x \to \ 0}\left(\frac{\sin\left(x\right)}{x}\right)^{2}+\lim_{x \to \ 0}\cos\left(x\right)\cdot\left(\frac{\cos\left(x\right)-1}{x^{2}}+\frac{1-\cos^{3}\left(2x\right)\cos^{2}\left(3x\right)}{x^{2}}\cdot\frac{1}{1+\sqrt{\cos\left(2x\right)}\sqrt[3]{\cos\left(3x\right)}}\cdot\frac{1}{\left(1+\cos\left(2x\right)\sqrt[3]{\cos^{2}\left(3x\right)+}\cos^{2}\left(2x\right)\sqrt[3]{\cos^{4}\left(3x\right)}\right)}\right)=$$$$=1+\left(-0.5+15\left(\frac{1}{2}\cdot\frac{1}{3}\right)\right)=3$$

the answer is true and I've checked that, but the solution I've used is not tricky as much as it expected, is there any better solution which does not use Taylor/Laurent series and also L'hopital's rule?

Answer 1

Abbreviate $\lim_{x\to0}f(x)=L$ as $f(x)\sim L$. Since $\frac{1-\cos x}{x^2}\sim \frac12$, $\frac{1-\cos nx}{x^2}=n^2\frac{1-\cos nx}{(nx)^2}\sim\frac12 n^2$ and$$\frac{1-\cos^{1/n}nx}{x^2}=\frac{1-\cos nx}{x^2}\frac{1}{\sum_{k=0}^{n-1}\cos^{k/n}nx}\sim\frac12n.$$If $\frac{1-f(x)}{x^2}\sim L$ and $\frac{1-g(x)}{x^2}\sim M$, $f(x)\sim 1$ and$$\frac{1-f(x)g(x)}{x^2}=\frac{1-f(x)}{x^2}+f(x)\frac{1-g(x)}{x^2}\sim L+1\cdot M=L+M.$$Hence$$\frac{1-\prod_{j=1}^3\cos^{1/j}jx}{x^2}=\sum_{j=1}^3\frac12j=3.$$

Answer 2

We can use that

$$\frac{1-\cos\left(x\right)\sqrt{\cos\left(2x\right)}\sqrt[3]{\cos\left(3x\right)}}{x^{2}}=$$

$$= \frac{\sqrt{\cos\left(2x\right)}\sqrt[3]{\cos\left(3x\right)}-\cos\left(x\right)\sqrt{\cos\left(2x\right)}\sqrt[3]{\cos\left(3x\right)}+1-\sqrt{\cos\left(2x\right)}\sqrt[3]{\cos\left(3x\right)}}{x^{2}}=$$

$$= \sqrt{\cos\left(2x\right)}\sqrt[3]{\cos\left(3x\right)}\frac{1-\cos\left(x\right)}{x^{2}} +\frac{1-\sqrt{\cos\left(2x\right)}\sqrt[3]{\cos\left(3x\right)}}{x^{2}}$$

and the first term tends to $\frac12$ then by the same trick

$$\frac{1-\sqrt{\cos\left(2x\right)}\sqrt[3]{\cos\left(3x\right)}}{x^{2}}=\frac{\sqrt{\cos\left(2x\right)}-\sqrt{\cos\left(2x\right)}\sqrt[3]{\cos\left(3x\right)}+1-\sqrt{\cos\left(2x\right)}}{x^{2}}=$$

$$=\sqrt{\cos\left(2x\right)}\frac{1-\sqrt[3]{\cos\left(3x\right)}}{x^{2}}+\frac{1-\sqrt{\cos\left(2x\right)}}{x^{2}}$$

and the first term tends to

$$\frac{1-\sqrt[3]{\cos\left(3x\right)}}{x^{2}}=9\frac{1-\cos\left(3x\right)}{(3x)^{2}}\frac1{1+(\cos(3x))^\frac13+(\cos^2(3x))^\frac23}\to \frac32$$

and the second

$$\frac{1-\sqrt{\cos\left(2x\right)}}{x^{2}}=4\frac{1-\cos\left(2x\right)}{(2x)^{2}} \frac1{1+\sqrt{\cos\left(2x\right)}}\to 1$$

Answer 3

Note that $$ \cos(x)=1-\tfrac12x^2+O\!\left(x^4\right)\tag1 $$ and $$ \left(1-x\right)^a=1-ax+O\!\left(x^2\right)\tag2 $$ and $$ (1+ax)(1+bx)=1+(a+b)x+O\!\left(x^2\right)\tag3 $$ Thus, $$ \begin{align} &1-\cos(x)\cos(2x)^{1/2}\cos(3x)^{1/3}\\ &=1-\left(1-\tfrac12x^2\right)\left(1-\tfrac12\cdot4x^2\right)^{1/2}\left(1-\tfrac12\cdot9x^2\right)^{1/3}+O\!\left(x^4\right)\tag4\\ &=1-\left(1-\tfrac12x^2\right)\left(1-x^2\right)\left(1-\tfrac32x^2\right)+O\!\left(x^4\right)\tag5\\ &=1-\left(1-3x^2\right)+O\!\left(x^4\right)\tag6\\ &=3x^2+O\!\left(x^4\right)\tag7 \end{align} $$ Explanation:
$(4)$: apply $(1)$
$(5)$: apply $(2)$
$(6)$: apply $(3)$
$(7)$: simplify

Therefore, $$ \begin{align} \lim_{x\to0}\frac{1-\cos(x)\cos(2x)^{1/2}\cos(3x)^{1/3}}{x^2} &=\lim_{x\to0}3+O\!\left(x^2\right)\\[6pt] &=3 \end{align} $$

Answer 4

$(I)$....$1=\lim_{y\to 0}(\sin y )/y=\lim_{y\to 0}(\sin^2 y)/y^2$ so $\sin^2 y =y^2(1+f(y))$ where $\lim_{y\to 0}f(y)=0.$ So we have $$\cos^2 y=1-\sin^2 y=1-y^2(1+f(y)).$$

$(IIa)$....$\cos^{12}x= (\cos^2 x)^6 =(1-x^2(1+f(x))\,)^6.$ Expanding this, since $f(x)\to 0$ as $x \to 0,$ we can write it as $$\cos^{12}x=1-6x^2 (1+g_1(x))$$ where $\lim_{x\to 0}g_1(x)=0.$

$(IIb)$....$\cos^6 2x=(\cos^2 2x)^3=(1-4x^2(1+f(2x))\,)^3.$ Expanding this, since $f(2x)\to 0$ as $x\to 0,$ we can write it as $$\cos^6 2x=1-12x^2(1+g_2(x))$$ where $\lim_{x\to 0}g_2(x)=0.$

$(IIc)$....$\cos^4 3x=(\cos^2 3x)^2 =(1-9x^2(1+f(3x))\,)^2.$ Expanding this , since $f(3x)\to 0$ as $x\to 0,$ we can write it as $$\cos^4 3x=1-18x^2(1+g_3(x))$$ where $\lim_{x\to 0}g_3(x)=0.$

$(IV)$.... Hence $$\cos^{12} x \cos^6 2x \cos^4 3x=(1-6x^2(1+g_1(x)) \cdot (1-12x^2(1+g_2(x)) \cdot (1-18x^2(1+g_3(x)).$$

Expanding this, since $g_1(x),g_2(x),$ and $g_3(x)$ all $\to 0 $ as $x\to 0,$ we can write this as $$\cos^{12}x \cos^6 2x \cos^4 3x=1-36 x^2(1+g_4(x))$$ where $\lim_{x\to 0}g_4(x)=0.$

$(V)$....For $|y|\le 1$ we have $(1-y)^{1/12}=1-(y/12)(1+h(y))$ where $\lim_{y\to 0}h(y)=0.$

$(VI)$...If $|x|$ is small enough we have $$1-(\cos x)(\cos 2x)^{1/2}(\cos 3x)^{1/3}=$$ $$=1-|(\cos x)(\cos 2x)^{1/2}(\cos 3x)^{1/3}|=$$ $$=1-(\cos^{12}x \cos^6 2x \cos^4 3x)^{1/12}=$$ $$=1-(1-36 x^2(1+g_4(x))\,)^{1/12}=$$ $$=1-(1-3x^2(1+j(x))\,)=$$ $$=3x^2(1+j(x))$$ where $j(x)=h(36x^2(1+g_4(x))\,)$ goes to $0$ as $x\to 0.$

Answer 5

You can easily handle the problem by noting that numerator is of the form $$1-abc$$ where each of $a, b, c$ tends to $1$. The same technique has been used in another answer to deal with a more complicated problem.

We can split numerator as $$1 - a +a(1-bc)$$ and then note that limit is split into sum of two limits $$\lim_{x\to 0}\frac{1-a}{x^2}+\lim_{x\to 0}\frac{1-bc}{x^2}$$ the second limit above can be handled again by same technique and thus the original limit is split as a sum of three limits.

The first of the limits $$\lim_{x\to 0}\frac{1-\cos x} {x^2}$$ evaluates to $1/2$ (pretty well known). The second limit is handled by writing it as $$\lim_{x\to 0}\frac{1-\sqrt {\cos 2x}} {1-\cos 2x} \cdot\frac{1-\cos 2x} {(2x)^2}\cdot 4=\frac{1}{2}\cdot\frac{1}{2}\cdot 4=1$$ And the third limit is also handled the same way $$\lim_{x\to 0}\frac{1-\sqrt[3]{\cos 3x}}{1-\cos 3x}\cdot\frac{1-\cos 3x}{(3x)^2}\cdot 9=\frac{1}{3}\cdot\frac{1}{2}\cdot 9=\frac{3}{2}$$ The desired limit is thus $(1/2) +1+(3/2) =3$.