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Question Number 84543 by niroj last updated on 14/Mar/20

Determine the value of a,b ,c so that _(x→0) ^(lim) (((a +b cos x) x−c sin x)/x^5 )=1

$$\boldsymbol{\mathrm{Determine}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\:,\boldsymbol{\mathrm{c}}\:\:\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{that}} \\ $$$$\:\:\underset{\mathrm{x}\rightarrow\mathrm{0}} {\overset{\mathrm{lim}} {\:}}\:\frac{\left(\boldsymbol{\mathrm{a}}\:+\boldsymbol{\mathrm{b}}\:\mathrm{cos}\:\mathrm{x}\right)\:\mathrm{x}−\boldsymbol{\mathrm{c}}\:\mathrm{sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{5}} }=\mathrm{1} \\ $$

Commented by mr W last updated on 14/Mar/20

a=120 b=60 c=180

$${a}=\mathrm{120} \\ $$$${b}=\mathrm{60} \\ $$$${c}=\mathrm{180} \\ $$

Commented by jagoll last updated on 14/Mar/20

lim_(x→0) (((a+b(1−(x^2 /2)+(x^4 /(24))+o(x^4 ))x−c(x−(x^3 /6)+(x^5 /(120))+o(x^5 )))/x^5 ) =1 lim_(x→0) ((ax+bx−((bx^3 )/2)+((bx^5 )/(24))−cx+((cx^3 )/6)−((cx^5 )/(120)))/x^5 )=1 lim_(x→0) (((a+b−c)x+(((c−3b)/6))x^3 +(((5b−c)/(120)))x^5 )/x^5 )=1

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

Commented by jagoll last updated on 14/Mar/20

c = 3b 5b−c =120 ⇒ 2b = 120 b = 60 ∧ c = 180 a+b−c = 0 ⇒ a = c−b = 120

$$\mathrm{c}\:=\:\mathrm{3b} \\ $$$$\mathrm{5b}−\mathrm{c}\:=\mathrm{120}\:\Rightarrow\:\mathrm{2b}\:=\:\mathrm{120}\: \\ $$$$\mathrm{b}\:=\:\mathrm{60}\:\wedge\:\mathrm{c}\:=\:\mathrm{180} \\ $$$$\mathrm{a}+\mathrm{b}−\mathrm{c}\:=\:\mathrm{0}\:\Rightarrow\:\mathrm{a}\:=\:\mathrm{c}−\mathrm{b}\:=\:\mathrm{120} \\ $$

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