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

$$\underset{x\to 0}{\lim }\frac{x(1+a\cos x)-b\sin x}{x^3}=1$$$$Findaandb.$$

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

$$\divideontimes \mathcal{JS}\divideontimes$$$$\underset{x\to 0}{\lim }\frac{x(1+a(1-\frac{x^2}{2}))-b(x-\frac{x^3}{6})}{x^3}=1$$$$\underset{x\to 0}{\lim }\frac{(1+a-b)x-\frac{{ax}^3}{2}+\frac{{bx}^3}{6}}{x^3}=1$$$$\to \{\begin{array}{l}1+a-b=0\to a-b=-1\\ \frac{b}{6}-\frac{a}{2}=1\to b-3a=6\end{array}$$$$\Rightarrow b-3(b-1)=6\Rightarrow -2b=3,b=-\frac{3}{2}$$$$anda=-\frac{3}{2}-1=-\frac{5}{2}$$

Commented by ajfour last updated on 10/Aug/20

$$Thanks,RightAnswer!$$

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

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

Commented by ajfour last updated on 10/Aug/20

$$Thankyou,Sir.$$

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