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

If lim_(x→0)  ((1−cos (1−cos x))/x^4 ) = (m/n) where m and n  are relative prime positive integer   then the sum of the digits (m^2 +n^2 ) equals

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

Answered by EDWIN88 last updated on 04/Feb/21

recall lim_(x→0) ((1−cos x)/x^2 ) = (1/2)  then lim_(x→0) ((1−cos (1−cos x))/x^4 )=lim_(x→0) ((1−cos (1−cos x))/((1−cos x)^2 )).(((1−cos x)^2 )/x^4 )  change variable : 1−cos x=u  =lim_(u→0) ((1−cos u)/u^2 )× lim_(x→0) (((1−cos x)/x^2 ))^2   = (1/2)×((1/2))^2 =(1/8) ≡ (m/n)  ⇔m^2 +n^2 = 65 ; sum of the digit=11

$${recall}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{x}}{{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${then}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}{{x}^{\mathrm{4}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}{\left(\mathrm{1}−\mathrm{cos}\:{x}\right)^{\mathrm{2}} }.\frac{\left(\mathrm{1}−\mathrm{cos}\:{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{4}} } \\ $$$${change}\:{variable}\::\:\mathrm{1}−\mathrm{cos}\:{x}={u} \\ $$$$=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{u}}{{u}^{\mathrm{2}} }×\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\mathrm{cos}\:{x}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{8}}\:\equiv\:\frac{{m}}{{n}} \\ $$$$\Leftrightarrow{m}^{\mathrm{2}} +{n}^{\mathrm{2}} =\:\mathrm{65}\:;\:{sum}\:{of}\:{the}\:{digit}=\mathrm{11} \\ $$

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