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as in topic, my task is to calculate $$\lim_{x \rightarrow 0}\left ( x^{-6}\cdot (1-\cos x^{\sin x})^2 \right )$$ I do the following: (assuming that de'Hospital is pointless here, as it seems to be) I write Taylor series for $\sin x, \cos x$ $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+o(x^7)$$ $$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+o(x^6)$$ I notice that denominator tells me to use sixth degree something and i get $$\frac{\cos x^{2 \cdot \sin x}- 2 \cdot (\cos x)^{\sin x} + 1}{x^6}$$ and my idea here was to use Bernuolli to $\cos x^{\sin x}$ so that $(1+x)^n\geq1+nx$ but it is going to be nasty and the expression $(...)\geq(...)$ does not tell anything about that limit. Any hints? Thank you in advance.

Julián Aguirre
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fdhd
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  • Use repeated L'Hospital rule? – Inceptio Mar 18 '13 at 09:39
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    It will be easier if you first figure out the limit of the square root. Apply L'Hospital rule once, you get: $$\lim_{x\to 0}\frac{1-\cos(x)^{\sin(x)}}{x^3} = \lim_{x\to 0} \left( -\cos(x)^{\sin(x)}\frac{\cos(x)\log(\cos(x)) - \frac{\sin(x)^2}{\cos(x)}}{3x^2}\right)$$ Some of the factors like $\cos(x)^{\sin(x)}$ converges to 1 as $x \to 0$, you don't need to worry around them. What's remain is manageable by Taylor expansion or L'Hospital rule. – achille hui Mar 18 '13 at 10:12

2 Answers2

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first find the first terms of the series expansion of $1-\cos(x)^{\sin(x)}$ without using the ones of sin and cos, you will find that$$1-\cos(x)^{\sin(x)}=\frac{x^3}{2}+\text{higer trems}$$ which implies $$\lim_{x\to 0}\frac{(1-\cos(x)^{\sin(x)})^2}{x^6}=\lim_{x\to 0}\frac{\frac{x^6}{4}+x^3(\text{higer terms})+\text{(higer terms)}^2}{x^6}=\frac{1}{4}$$ since the degree of higer terms is larger than 3.

Mikasa
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$$\lim_{x \rightarrow 0}\left ( x^{-6}\cdot (1-\cos x^{\sin x})^2 \right )$$

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

Now, Putting $y=\sin x$ in the last limit, as $x\to 0,y=\sin x\to 0$ and $\cos x^{\sin x}=(\cos^2x)^{\frac {\sin x}2}=(1-y^2)^\frac y2$

$$\lim_{x\to 0}\frac{1-\cos x^{\sin x}}{\sin^3x}$$ $$=\lim_{y\to 0}\frac{1-(1-y^2)^{\frac y2}}{y^3}$$

$$=\lim_{y\to 0}\frac{1-\left(1+(-y^2)\cdot \frac y2+(-y^2)^2\cdot\frac{\frac y2\left(\frac y2-1\right)}{2!}+\cdots\right)}{y^3}$$

$$=\lim_{y\to 0}\{\frac12+O(y^2)\}\text { as }y\ne0 \text { as } y\to 0$$

$$=\frac12$$

lab bhattacharjee
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