The values of a and b such that mathop {lim } | vedclass

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(C) Given $\mathop {\lim }\limits_{x 	o 0} \frac{{x(1 + a\cos x) - b\sin x}}{{{x^3}}} = 1$.Using the Taylor series expansion for $\cos x = 1 - \frac{

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$a = -\frac{5}{2}, b = -\frac{3}{2}$

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(C) Given $\mathop {\lim }\limits_{x 	o 0} \frac{{x(1 + a\cos x) - b\sin x}}{{{x^3}}} = 1$.Using the Taylor series expansion for $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$ and $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$:$\mathop {\lim }\limits_{x 	o 0} \frac{{x(1 + a(1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots)) - b(x - \frac{x^3}{6} + \frac{x^5}{120} - \dots)}}{{{x^3}}} = 1$$\mathop {\lim }\limits_{x 	o 0} \frac{{x(1 + a - b) + x^3(\frac{b}{6} - \frac{a}{2}) + x^5(\frac{a}{24} - \frac{b}{120}) + \dots}}{{{x^3}}} = 1$For the limit to exist and equal $1$,the coefficient of $x$ must be $0$:$1 + a - b = 0 \implies b - a = 1$And the coefficient of $x^3$ must be $1$:$\frac{b}{6} - \frac{a}{2} = 1 \implies b - 3a = 6$Subtracting the equations: $(b - 3a) - (b - a) = 6 - 1 \implies -2a = 5 \implies a = -\frac{5}{2}$.Substituting $a$ into $b - a = 1$: $b - (-\frac{5}{2}) = 1 \implies b = 1 - \frac{5}{2} = -\frac{3}{2}$.Thus,$a = -\frac{5}{2}$ and $b = -\frac{3}{2}$.

The values of $a$ and $b$ such that $\mathop {\lim }\limits_{x \to 0} \frac{{x(1 + a\cos x) - b\sin x}}{{{x^3}}} = 1$ are:

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