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Question Number 38893 by gunawan last updated on 01/Jul/18

lim_(x→0) ((x(a+b cos x)−c sin x)/x^5 )=1 find the value a,b, and c

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

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18

lim_(x→0) ((ax+bx(1−(x^2 /(2!))+(x^4 /(4!))+...)−c((x/1)−(x^3 /(3!))+(x^5 /(5!))+..))/x^5 ) =lim_(x→0) ((x(a+b−c)+b(((−x^3 )/(2!))+(x^5 /(4!))+..)−c(−(x^3 /(3!))+(x^5 /(5!))..))/x^5 ) ((lim)/(x→0))(((a+b−c)+b(−((3x^2 )/(2!))+((5x^4 )/(4!))+..)−c(((−3x^2 )/(3!))+((5x^4 )/(5!))+..)/(5x^4 )) so for((0/0)) form a+b−c=0.....eqn 1 =lim_(x→0) ((b(−(3/(2!))+((5x^2 )/(4!))+..)−c(((−3)/(3!))+((5x^2 )/(5!))+..))/(5x^2 )) b(−(3/(2!)))+c((3/(3!)))=0 for((0/0)) form (c/2)−((3b)/2)=0 c=3b lim_(x→0) ((b(((5x^2 )/(4!))+..)−c(((5x^2 )/(5!))+..))/(5x^2 ))=1 (b/(4!))−(c/(5!))=1 (b/(24))−(c/(120))=1 (b/(24))−((3b)/(120))=1 (b/(24))−(b/(40))=1 so ((5b−3b)/(3×8×5))=1 2b=3×8×5 b=60 c=3×60=180 a+b−c=0 a=c−b a=180−60=120

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ax}+{bx}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+...\right)−{c}\left(\frac{{x}}{\mathrm{1}}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}+..\right)}{{x}^{\mathrm{5}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\left({a}+{b}−{c}\right)+{b}\left(\frac{−{x}^{\mathrm{3}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{4}!}+..\right)−{c}\left(−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}..\right)}{{x}^{\mathrm{5}} } \\ $$$$\frac{{lim}}{{x}\rightarrow\mathrm{0}}\frac{\left({a}+{b}−{c}\right)+{b}\left(−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{5}{x}^{\mathrm{4}} }{\mathrm{4}!}+..\right)−{c}\left(\frac{−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{3}!}+\frac{\mathrm{5}{x}^{\mathrm{4}} }{\mathrm{5}!}+..\right.}{\mathrm{5}{x}^{\mathrm{4}} } \\ $$$${so}\:{for}\left(\frac{\mathrm{0}}{\mathrm{0}}\right)\:{form}\:{a}+{b}−{c}=\mathrm{0}.....{eqn}\:\mathrm{1} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{b}\left(−\frac{\mathrm{3}}{\mathrm{2}!}+\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{4}!}+..\right)−{c}\left(\frac{−\mathrm{3}}{\mathrm{3}!}+\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{5}!}+..\right)}{\mathrm{5}{x}^{\mathrm{2}} } \\ $$$${b}\left(−\frac{\mathrm{3}}{\mathrm{2}!}\right)+{c}\left(\frac{\mathrm{3}}{\mathrm{3}!}\right)=\mathrm{0}\:\:\:{for}\left(\frac{\mathrm{0}}{\mathrm{0}}\right)\:{form} \\ $$$$\frac{{c}}{\mathrm{2}}−\frac{\mathrm{3}{b}}{\mathrm{2}}=\mathrm{0}\:\:\:{c}=\mathrm{3}{b} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{b}\left(\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{4}!}+..\right)−{c}\left(\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{5}!}+..\right)}{\mathrm{5}{x}^{\mathrm{2}} }=\mathrm{1}\:\: \\ $$$$\frac{{b}}{\mathrm{4}!}−\frac{{c}}{\mathrm{5}!}=\mathrm{1}\:\:\:\:\:\frac{{b}}{\mathrm{24}}−\frac{{c}}{\mathrm{120}}=\mathrm{1} \\ $$$$\frac{{b}}{\mathrm{24}}−\frac{\mathrm{3}{b}}{\mathrm{120}}=\mathrm{1}\:\:\: \\ $$$$\frac{{b}}{\mathrm{24}}−\frac{{b}}{\mathrm{40}}=\mathrm{1}\:\:\:\:\:{so}\:\frac{\mathrm{5}{b}−\mathrm{3}{b}}{\mathrm{3}×\mathrm{8}×\mathrm{5}}=\mathrm{1}\:\: \\ $$$$\mathrm{2}{b}=\mathrm{3}×\mathrm{8}×\mathrm{5}\:\:\:\:{b}=\mathrm{60}\:\: \\ $$$${c}=\mathrm{3}×\mathrm{60}=\mathrm{180} \\ $$$${a}+{b}−{c}=\mathrm{0}\:\:\:{a}={c}−{b} \\ $$$${a}=\mathrm{180}−\mathrm{60}=\mathrm{120} \\ $$$$ \\ $$

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