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Question Number 107377 by ajfour last updated on 10/Aug/20

lim_(x→0) ((x(1+acos x)−bsin x)/x^3 ) = 1 Find a and b .

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\left(\mathrm{1}+{a}\mathrm{cos}\:{x}\right)−{b}\mathrm{sin}\:{x}}{{x}^{\mathrm{3}} }\:=\:\mathrm{1} \\ $$$${Find}\:\:{a}\:{and}\:{b}\:. \\ $$

Answered by john santu last updated on 10/Aug/20

⋇JS⋇ lim_(x→0) ((x(1+a(1−(x^2 /2)))−b(x−(x^3 /6)))/x^3 )=1 lim_(x→0) (((1+a−b)x−((ax^3 )/2)+((bx^3 )/6))/x^3 )=1 → { ((1+a−b=0→a−b=−1)),(((b/6)−(a/2)=1→b−3a=6)) :} ⇒b−3(b−1)=6 ⇒−2b=3 ,b=−(3/2) and a=−(3/2)−1=−(5/2)

$$\:\:\:\:\divideontimes\mathcal{JS}\divideontimes \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\left(\mathrm{1}+{a}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\right)−{b}\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)}{{x}^{\mathrm{3}} }=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+{a}−{b}\right){x}−\frac{{ax}^{\mathrm{3}} }{\mathrm{2}}+\frac{{bx}^{\mathrm{3}} }{\mathrm{6}}}{{x}^{\mathrm{3}} }=\mathrm{1} \\ $$$$\rightarrow\begin{cases}{\mathrm{1}+{a}−{b}=\mathrm{0}\rightarrow{a}−{b}=−\mathrm{1}}\\{\frac{{b}}{\mathrm{6}}−\frac{{a}}{\mathrm{2}}=\mathrm{1}\rightarrow{b}−\mathrm{3}{a}=\mathrm{6}}\end{cases} \\ $$$$\Rightarrow{b}−\mathrm{3}\left({b}−\mathrm{1}\right)=\mathrm{6}\:\Rightarrow−\mathrm{2}{b}=\mathrm{3}\:,{b}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${and}\:{a}=−\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}=−\frac{\mathrm{5}}{\mathrm{2}} \\ $$

Commented by ajfour last updated on 10/Aug/20

Thanks, Right Answer!

$${Thanks},\:{Right}\:{Answer}! \\ $$

Answered by Dwaipayan Shikari last updated on 10/Aug/20

lim_(x→0) ((−axsinx+(1+acosx)−bcosx)/(3x^2 ))=1 lim_(x→0) ((−asinx−axcosx−asinx+bsinx)/(6x))=1 lim_(x→0) ((−2ax−axcosx+bx)/(6x))=1 −2a−acosx+b=6 b−3a=6→(1) Or ((x(1+acosx)−bsinx)/x^3 )=1 ((x+axcosx−bx)/x^3 )=1 x+axcosx−bx=x^3 1+a−b=x^2 a−b=−1→(2) { ((b=−(3/2))),((a=−(5/2))) :}

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{axsinx}+\left(\mathrm{1}+\mathrm{acosx}\right)−\mathrm{bcosx}}{\mathrm{3x}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{asinx}−\mathrm{axcosx}−\mathrm{asinx}+\mathrm{bsinx}}{\mathrm{6x}}=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2ax}−\mathrm{axcosx}+\mathrm{bx}}{\mathrm{6x}}=\mathrm{1} \\ $$$$−\mathrm{2a}−\mathrm{acosx}+\mathrm{b}=\mathrm{6} \\ $$$$\mathrm{b}−\mathrm{3a}=\mathrm{6}\rightarrow\left(\mathrm{1}\right) \\ $$$$\mathrm{Or} \\ $$$$\frac{\mathrm{x}\left(\mathrm{1}+\mathrm{acosx}\right)−\mathrm{bsinx}}{\mathrm{x}^{\mathrm{3}} }=\mathrm{1} \\ $$$$\frac{\mathrm{x}+\mathrm{axcosx}−\mathrm{bx}}{\mathrm{x}^{\mathrm{3}} }=\mathrm{1} \\ $$$$\mathrm{x}+\mathrm{axcosx}−\mathrm{bx}=\mathrm{x}^{\mathrm{3}} \\ $$$$\mathrm{1}+\mathrm{a}−\mathrm{b}=\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{a}−\mathrm{b}=−\mathrm{1}\rightarrow\left(\mathrm{2}\right) \\ $$$$\begin{cases}{\mathrm{b}=−\frac{\mathrm{3}}{\mathrm{2}}}\\{\mathrm{a}=−\frac{\mathrm{5}}{\mathrm{2}}}\end{cases} \\ $$$$ \\ $$$$ \\ $$

Commented by ajfour last updated on 10/Aug/20

Thank you, Sir.

$${Thank}\:{you},\:{Sir}. \\ $$

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