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How can I calculate following limit without L'Hôpital's rule

$$\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$$

I tried L'Hôpital's rule and I found the result $2$.

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  • $\begingroup$ "I tried L'Hopital's rule and I found the result $2$." What's the problem then? $\endgroup$ – infinitylord Dec 16 '17 at 8:21
  • $\begingroup$ Thank you for fixed question. $\endgroup$ – Soru Dec 16 '17 at 8:23
  • $\begingroup$ I'm also curious. Did you not trust the answer you got with L'hopital for some reason? Did you the think this answer is different if you don't use L'hopital? $\endgroup$ – fleablood Dec 16 '17 at 8:26
  • $\begingroup$ Lopital rule is so hard this for me. $\endgroup$ – Soru Dec 16 '17 at 8:28
  • $\begingroup$ @Soru If you are ok, you can set as solved. Thanks! $\endgroup$ – user Dec 17 '17 at 8:02
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This is essentially the same as MyGlasses's answer (which I overlooked at first), just presented in a different order.

First, a little preliminary:

$$\begin{align} {1-\sqrt[n]{\cos nx}\over x^2} &=\left(1-\sqrt[n]{\cos nx}\over1-\cos nx\right)\left(1-\cos nx\over x^2\right)\\ &=\left(1-\sqrt[n]{\cos nx}\over1-\cos nx\right)\left(1-\cos^2nx\over (nx)^2\right)\left(n^2\over1+\cos nx\right)\\ &=\left(1-\sqrt[n]{\cos nx}\over1-\cos nx\right)\left(\sin nx\over nx\right)^2\left(n^2\over1+\cos nx\right)\\ &=\left(1-u\over1-u^n\right)\left(\sin\theta\over\theta\right)^2\left(n^2\over1+\cos nx\right)\\ &=\left(1\over1+u+\cdots+u^{n-1} \right)\left(\sin\theta\over\theta\right)^2\left(n^2\over1+\cos nx\right)\\ &\to\left(1\over1+1+\cdots+1 \right)(1)^2\left(n^2\over1+1 \right)={n\over2} \end{align}$$

where we have $u=\sqrt[n]{\cos nx}\to1$ as $\theta=nx\to0$.

Now, as in MyGlasses's answer, we have

$$\begin{align}{\cos x-\sqrt{\cos2x}\sqrt[3]{\cos3x}\over x^2} &={\sqrt{\cos2x}(1-\sqrt[3]{\cos3x})+(1-\sqrt{\cos2x})-(1-\cos x)\over x^2}\\\\ &=\sqrt{\cos2x}\left(1-\sqrt[3]{\cos3x}\over x^2 \right)+\left(1-\sqrt{\cos2x}\over x^2\right)-\left(1-\cos x\over x^2\right)\\\\ &\to(1)\left(3\over2\right)+\left(2\over2\right)-\left(1\over2\right)=2 \end{align}$$

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Write $$\lim_{x\to0}\dfrac{\cos x-1+1-\sqrt{\cos 2x}+\sqrt{\cos 2x}-\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}=$$ $$\lim_{x\to0}\dfrac{\cos x-1}{x^2}+\lim_{x\to0}\dfrac{1-\sqrt{\cos 2x}}{x^2}+\lim_{x\to0}\sqrt{\cos 2x}\lim_{x\to0}\dfrac{1-\sqrt[3]{\cos 3x}}{x^2}$$ and we have \begin{align} &\lim_{x\to0}\dfrac{\cos x-1}{x^2}=\lim_{x\to0}\dfrac{-2\sin^2\frac{x}{2}}{x^2}=-\dfrac{1}{2} \\ &\lim_{x\to0}\dfrac{1-\sqrt{\cos 2x}}{x^2}=\lim_{x\to0}\dfrac{1-\cos 2x}{x^2(1+\sqrt{\cos 2x})}=\lim_{x\to0}\dfrac{2\sin^2x}{x^2(1+\sqrt{\cos 2x})}=1 \\ &\lim_{x\to0}\sqrt{\cos 2x}=1 \\ &\lim_{x\to0}\dfrac{1-\sqrt[3]{\cos 3x}}{x^2}=\lim_{x\to0}\dfrac{1-\cos3x}{x^2(1+\sqrt[3]{\cos 3x}+\sqrt[3]{\cos^23x})}=\dfrac{3}{2} \end{align} the answer is $2$.

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$$\cos x=1-\frac{x^2}2+O(x^4),$$ $$\cos nx=1-\frac{n^2x^2}2+O(x^4),$$ $$(\cos nx)^{1/n}=1-\frac{nx^2}2+O(x^4),$$ $$(\cos 2x)^{1/2}(\cos 3x)^{1/3} =\left(1-\frac{2x^2}{2}+O(x^4)\right)\left(1-\frac{3x^2}{2}+O(x^4)\right) =1-\frac{5x^2}{2}+O(x^4)$$ and so $$\cos x-(\cos 2x)^{1/2}(\cos 3x)^{1/3}=2x^2+O(x^4)$$ so the limit you seek is $2$.

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  • $\begingroup$ What is $O$ (x)..I dont understood :( $\endgroup$ – Soru Dec 16 '17 at 8:26
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    $\begingroup$ @Soru Big O notation: very useful to master en.wikipedia.org/wiki/Big_O_notation $\endgroup$ – Lord Shark the Unknown Dec 16 '17 at 8:28
  • $\begingroup$ Thank you..I read. so interesting. I learned new thing. $\endgroup$ – Soru Dec 16 '17 at 8:37
  • $\begingroup$ @Soru: See here for precise versions of this technique, and the linked examples to see how powerful it is. Also see here for some motivation and elementary ways to obtain simple asymptotic approximations for square/cube-roots. =) $\endgroup$ – user21820 Dec 16 '17 at 9:58
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$$\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}\sqrt[3] {\cos 3x}}{x^2}=$$ $$=\lim_{x\rightarrow0}\frac{\cos^6x-\cos^32x\cos^23x}{6x^2}=$$ $$=\lim_{x\rightarrow0}\frac{\cos^2x(1-\cos^2x)(128\cos^8x-256\cos^6x+200\cos^4x-69\cos^2x+9)}{6x^2}=$$ $$=\frac{128-256+200-69+9}{6}=2.$$ I used $$x^6-y^6=(x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^5+y^5),$$ $$\lim_{x\rightarrow0}\frac{\sin{x}}{x}=1$$ and $$\lim_{x\rightarrow0}\cos{x}=1.$$

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Hint:

$$\cos x-(\cos2x)^{1/2}(\cos3x)^{1/3}=1-\cos x-[1-(\cos2x)^{1/2}(\cos3x)^{1/3}]$$

Now $\lim_{x\to0}\dfrac{1-\cos x}{x^2}=\cdots=\dfrac12$

On rationalization using $a^6-b^6=(a-b)(\cdots),$

$$1-(\cos2x)^{1/2}(\cos3x)^{1/3}=\dfrac{1-\cos^32x\cos^23x}{\sum_{r=0}^5[(\cos2x)^{1/2}(\cos3x)^{1/3}]^r}$$

$\lim_{x\to0}\sum_{r=0}^5[(\cos2x)^{1/2}(\cos3x)^{1/3}]^r=\sum_{r=0}^51=?$

Finally, $1-\cos^32x\cos^23x=1-(1-2\sin^2x)^3(1-\sin^23x)$

$\approx1-(1-2x^2)^3(1-9x^2)=x^2(6+9)+O(x^4)$

as $\lim_{x\to0}\dfrac{\sin mx}{mx}=1$

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By power series we need the following two:

$$\cos x=1-\frac{x^2}{2}+o(x^2)$$

$$(1+x)^n=1+nx+o(x)$$

Thus

$$\sqrt {\cos 2x}=\left( 1-\frac{4x^2}{2}+o(x^2)\right)^\frac12=1-x^2+o(x^2)$$

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

Thus

$$\sqrt {\cos 2x}×\sqrt[3] {\cos 3x}=\left(1-x^2+o(x^2)\right) \left(1-\frac{3x^2}{2}+o(x^2)\right)=1-\frac{5x^2}{2}+o(x^2)$$

And then:

$$\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}= \frac{1-\frac{x^2}{2}-1+\frac{5x^2}{2}+o(x^2)}{x^2}=\frac{2x^2+o(x^2)}{x^2}=2+o(1)\to2$$

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    $\begingroup$ You have $(1+x)^{n}=1+nx+o(x)$ and not $o(x^2)$. Apart from this everything looks fine. $\endgroup$ – Paramanand Singh Dec 16 '17 at 12:35
  • $\begingroup$ ops of course, thanks @ParamanandSingh :) $\endgroup$ – user Dec 16 '17 at 12:36

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