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.

  • $\begingroup$ Use repeated L'Hospital rule? $\endgroup$ – Inceptio Mar 18 '13 at 9:39
  • 1
    $\begingroup$ 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. $\endgroup$ – achille hui Mar 18 '13 at 10:12

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

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

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

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

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

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



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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.