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Question Number 131369 by liberty last updated on 04/Feb/21

$$\mathrm{If}\underset{x\to 0}{\lim }\frac{1-\cos (1-\cos \mathrm{x})}{{\mathrm{x}}^4}=\frac{\mathrm{m}}{\mathrm{n}}\mathrm{where}\mathrm{m}\mathrm{and}\mathrm{n}$$$$\mathrm{are}\mathrm{relative}\mathrm{prime}\mathrm{positive}\mathrm{integer}$$$$\mathrm{then}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{digits}({\mathrm{m}}^2+{\mathrm{n}}^2)\mathrm{equals}$$

Answered by EDWIN88 last updated on 04/Feb/21

$$recall\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}=\frac{1}{2}$$$$then\underset{x\to 0}{\lim }\frac{1-\cos (1-\cos x)}{x^4}=\underset{x\to 0}{\lim }\frac{1-\cos (1-\cos x)}{{(1-\cos x)}^2}.\frac{{(1-\cos x)}^2}{x^4}$$$$changevariable:1-\cos x=u$$$$=\underset{u\to 0}{\lim }\frac{1-\cos u}{u^2}\times \underset{x\to 0}{\lim }{\left(\frac{1-\cos x}{x^2}\right)}^2$$$$=\frac{1}{2}\times {\left(\frac{1}{2}\right)}^2=\frac{1}{8}\equiv \frac{m}{n}$$$$\iff m^2+n^2=65;sumofthedigit=11$$

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