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Question Number 105410 by john santu last updated on 28/Jul/20

lim_(x→0) ((x(1+a cos x)−bsin x)/x^5 ) = 1 find a & b

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\left(\mathrm{1}+{a}\:\mathrm{cos}\:{x}\right)−{b}\mathrm{sin}\:{x}}{{x}^{\mathrm{5}} }\:=\:\mathrm{1} \\ $$ $${find}\:{a}\:\&\:{b}\: \\ $$

Answered by bobhans last updated on 29/Jul/20

lim_(x→0) ((x(1+a(1−(1/2)x^2 +(1/(24))x^4 ))−b(x−(1/6)x^3 +(x^5 /(120))))/x^5 )=1 lim_(x→0) ((x(1+a−(1/2)ax^2 +(1/(24))ax^4 )−(bx−(1/6)bx^3 +(b/(120))x^5 ))/x^5 )=1 lim_(x→0) (((1+a−b)x+((1/6)b−(1/2)a)x^3 +((1/(24))a−(1/(120))b)x^5 )/x^5 )=1 (1)1+a−b=0 ; a−b=−1 (2)(1/6)b−(1/2)a=0 ; b=3a (3)(1/(24))a−(1/(120))b=1;a=(1/5)b the question inconsistent

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\left(\mathrm{1}+{a}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{24}}{x}^{\mathrm{4}} \right)\right)−{b}\left({x}−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} +\frac{{x}^{\mathrm{5}} }{\mathrm{120}}\right)}{{x}^{\mathrm{5}} }=\mathrm{1} \\ $$ $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\left(\mathrm{1}+{a}−\frac{\mathrm{1}}{\mathrm{2}}{ax}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{24}}{ax}^{\mathrm{4}} \right)−\left({bx}−\frac{\mathrm{1}}{\mathrm{6}}{bx}^{\mathrm{3}} +\frac{{b}}{\mathrm{120}}{x}^{\mathrm{5}} \right)}{{x}^{\mathrm{5}} }=\mathrm{1} \\ $$ $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+{a}−{b}\right){x}+\left(\frac{\mathrm{1}}{\mathrm{6}}{b}−\frac{\mathrm{1}}{\mathrm{2}}{a}\right){x}^{\mathrm{3}} +\left(\frac{\mathrm{1}}{\mathrm{24}}{a}−\frac{\mathrm{1}}{\mathrm{120}}{b}\right){x}^{\mathrm{5}} }{{x}^{\mathrm{5}} }=\mathrm{1} \\ $$ $$\left(\mathrm{1}\right)\mathrm{1}+{a}−{b}=\mathrm{0}\:;\:{a}−{b}=−\mathrm{1} \\ $$ $$\left(\mathrm{2}\right)\frac{\mathrm{1}}{\mathrm{6}}{b}−\frac{\mathrm{1}}{\mathrm{2}}{a}=\mathrm{0}\:;\:{b}=\mathrm{3}{a} \\ $$ $$\left(\mathrm{3}\right)\frac{\mathrm{1}}{\mathrm{24}}{a}−\frac{\mathrm{1}}{\mathrm{120}}{b}=\mathrm{1};{a}=\frac{\mathrm{1}}{\mathrm{5}}{b} \\ $$ $${the}\:{question}\:{inconsistent} \\ $$

Commented by1549442205PVT last updated on 29/Jul/20

I think that in the question need to give three unknown

$$\mathrm{I}\:\mathrm{think}\:\mathrm{that}\:\mathrm{in}\:\mathrm{the}\:\mathrm{question}\:\mathrm{need}\:\mathrm{to}\:\mathrm{give}\: \\ $$ $$\mathrm{three}\:\mathrm{unknown}\: \\ $$

Answered by Dwaipayan Shikari last updated on 29/Jul/20

lim_(x→0) ((x(1+acosx)−bsinx)/x^5 )=((xa(−sinx)+acosx−bcosx)/(5x^4 )) =((xa(−cosx)+a(−sinx)+a(−sinx)+bsinx)/(20x^3 )) =((xasinx+a(−cosx)+a(−cosx)+a(−cosx)+bcosx)/(60x^2 )) =((xacosx+asinx+3asinx−bsinx)/(120x))or((xa(−sinx)+acosx+4acosx−bcosx)/(120)) =((acosx)/(120))+((ax)/(120x))+((3ax)/(120x))−((bx)/(120x)) or ((5acosx−bcosx)/(120)) =(a/(24))−(b/(120)) ⇒(a/(24))−(b/(120))=1 5a−b=120→(1) lim_(x→0) ((x(1+acosx)−bsinx)/x^5 )=1 1+acosx−b=0 1+a=b→(2) 5a−1−a=120 a=((121)/4) b=((125)/4)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\left(\mathrm{1}+{acosx}\right)−{bsinx}}{{x}^{\mathrm{5}} }=\frac{{xa}\left(−{sinx}\right)+{acosx}−{bcosx}}{\mathrm{5}{x}^{\mathrm{4}} } \\ $$ $$=\frac{{xa}\left(−{cosx}\right)+{a}\left(−{sinx}\right)+{a}\left(−{sinx}\right)+{bsinx}}{\mathrm{20}{x}^{\mathrm{3}} } \\ $$ $$=\frac{{xasinx}+{a}\left(−{cosx}\right)+{a}\left(−{cosx}\right)+{a}\left(−{cosx}\right)+{bcosx}}{\mathrm{60}{x}^{\mathrm{2}} } \\ $$ $$=\frac{{xacosx}+{asinx}+\mathrm{3}{asinx}−{bsinx}}{\mathrm{120}{x}}{or}\frac{{xa}\left(−{sinx}\right)+{acosx}+\mathrm{4}{acosx}−{bcosx}}{\mathrm{120}} \\ $$ $$=\frac{{acosx}}{\mathrm{120}}+\frac{{ax}}{\mathrm{120}{x}}+\frac{\mathrm{3}{ax}}{\mathrm{120}{x}}−\frac{{bx}}{\mathrm{120}{x}}\:\:\:\:\:{or}\:\:\frac{\mathrm{5}{acosx}−{bcosx}}{\mathrm{120}} \\ $$ $$=\frac{{a}}{\mathrm{24}}−\frac{{b}}{\mathrm{120}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ $$\Rightarrow\frac{{a}}{\mathrm{24}}−\frac{{b}}{\mathrm{120}}=\mathrm{1} \\ $$ $$\mathrm{5}{a}−{b}=\mathrm{120}\rightarrow\left(\mathrm{1}\right) \\ $$ $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\left(\mathrm{1}+{acosx}\right)−{bsinx}}{{x}^{\mathrm{5}} }=\mathrm{1} \\ $$ $$\mathrm{1}+{acosx}−{b}=\mathrm{0} \\ $$ $$\mathrm{1}+{a}={b}\rightarrow\left(\mathrm{2}\right) \\ $$ $$\mathrm{5}{a}−\mathrm{1}−{a}=\mathrm{120} \\ $$ $${a}=\frac{\mathrm{121}}{\mathrm{4}} \\ $$ $${b}=\frac{\mathrm{125}}{\mathrm{4}} \\ $$ $$ \\ $$

Commented by1549442205PVT last updated on 30/Jul/20

♮ lim((x(1+acosx)−bsinx)/x^5 )=1 1+acosx−b=0 ε(1) I think that ocurred a mistake at this place.It is only possible 1+acosx−b=0⇒ lim((x(1+acosx)−bsinx)/x^5 )=1 and it is impossible ⇒lim((x(1+acosx)−bsinx)/x^5 )=1⇒1+acosx−b=0 it is also like as 1+acosx−kb=0 ⇒lim((x(1+acosx)−bsinx)/x^5 )=1 If (1) is true then 1+acosx−kb=0 is also true.Sir should see once again sir′argument

$$\natural\:\mathrm{lim}\frac{{x}\left(\mathrm{1}+{acosx}\right)−{bsinx}}{{x}^{\mathrm{5}} }=\mathrm{1} \\ $$ $$\mathrm{1}+{acosx}−{b}=\mathrm{0}\:\:\varepsilon\left(\mathrm{1}\right) \\ $$ $$\mathrm{I}\:\mathrm{think}\:\mathrm{that}\:\mathrm{ocurred}\:\mathrm{a}\:\mathrm{mistake}\:\mathrm{at}\:\mathrm{this} \\ $$ $$\mathrm{place}.\mathrm{It}\:\mathrm{is}\:\mathrm{only}\:\mathrm{possible}\: \\ $$ $$\mathrm{1}+{acosx}−{b}=\mathrm{0}\Rightarrow\:\mathrm{lim}\frac{{x}\left(\mathrm{1}+{acosx}\right)−{bsinx}}{{x}^{\mathrm{5}} }=\mathrm{1} \\ $$ $$\mathrm{and}\:\mathrm{it}\:\mathrm{is}\:\mathrm{impossible} \\ $$ $$\Rightarrow\mathrm{lim}\frac{{x}\left(\mathrm{1}+{acosx}\right)−{bsinx}}{{x}^{\mathrm{5}} }=\mathrm{1}\Rightarrow\mathrm{1}+{acosx}−{b}=\mathrm{0} \\ $$ $$\mathrm{it}\:\mathrm{is}\:\mathrm{also}\:\mathrm{like}\:\mathrm{as}\:\mathrm{1}+\mathrm{acosx}−\mathrm{kb}=\mathrm{0} \\ $$ $$\Rightarrow\mathrm{lim}\frac{{x}\left(\mathrm{1}+{acosx}\right)−{bsinx}}{{x}^{\mathrm{5}} }=\mathrm{1} \\ $$ $$\mathrm{If}\:\left(\mathrm{1}\right)\:\mathrm{is}\:\mathrm{true}\:\mathrm{then}\:\:\mathrm{1}+\mathrm{acosx}−\mathrm{kb}=\mathrm{0} \\ $$ $$\mathrm{is}\:\mathrm{also}\:\mathrm{true}.\mathrm{Sir}\:\mathrm{should}\:\mathrm{see}\:\mathrm{once}\:\mathrm{again} \\ $$ $$\mathrm{sir}'\mathrm{argument} \\ $$

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