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Question Number 103255 by Study last updated on 13/Jul/20

$$li\underset{x\to 0}{m}{\left(\frac{{sin}^6\sqrt{2}x}{{tan}^7\frac{x}{2}}\right)}^{{cos}^{12}x}=???$$

Commented by Dwaipayan Shikari last updated on 13/Jul/20

$$\mathrm{May}\mathrm{be}\mathrm{infinity}$$

Answered by OlafThorendsen last updated on 13/Jul/20

$$\sin u\underset{0}{\sim }u$$$$\tan u\underset{0}{\sim }u$$$$\mathrm{Then}\frac{{\sin }^6\sqrt{2}x}{{\tan }^7\frac{x}{2}}\underset{0}{\sim }\frac{{\left(\sqrt{2}x\right)}^6}{{\left(\frac{x}{2}\right)}^7}=\frac{2^3{x}^6}{\frac{x^7}{2^7}}=\frac{1024}{x}$$$$\mathrm{And}\cos x\underset{0}{\sim }1-\frac{x^2}{2}\Rightarrow {\cos }^{12}x\underset{0}{\sim }1-6x^2$$$$\underset{x\to 0}{\lim }{\left(\frac{{\sin }^6\sqrt{2}x}{{\tan }^7\frac{x}{2}}\right)}^{{\cos }^{12}x}=\underset{x\to 0}{\lim }{\left(\frac{1024}{x}\right)}^{1-6x^2}$$$$\underset{x\to 0}{\lim }{\left(\frac{{\sin }^6\sqrt{2}x}{{\tan }^7\frac{x}{2}}\right)}^{{\cos }^{12}x}=\{\begin{array}{l}-\mathrm{\infty }\mathrm{if}x<0\\ +\mathrm{\infty }\mathrm{if}x>0\end{array}$$$$$$

Commented by Study last updated on 14/Jul/20

$$whycosx\sim 1-\frac{x^2}{2}\Rightarrow {cos}^{12}x\sim 1-6x^2$$$$$$$$$$$$$$

Commented by OlafThorendsen last updated on 14/Jul/20

$$\cos x\sim 1-\frac{x^2}{2}$$$${(1+u)}^{\alpha }\sim 1+\alpha u$$$$\alpha =12\mathrm{and}u=-\frac{x^2}{2}$$$$\mathrm{Then}{\cos }^{12}x\sim {(1-\frac{x^2}{2})}^{12}\sim 1-6x^2$$

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