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$\lim\limits_{x\to 0}\large\frac{x(1+a\cos x)-b\sin x}{x^3}$$=1$ then $a,b$ are

$\begin{array}{1 1}(A)\;\large\frac{1}{2},\frac{-3}{2}&(B)\;\large\frac{5}{2},\frac{3}{2}\\(C)\;\large\frac{-5}{2},\frac{-3}{2}&(D)\;\text{None of these}\end{array}$

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Using expansions of $\sin x$ and $\cos x$.
Given limit=$\lim\limits_{x\to 0}\large\frac{x+ax(1-\Large\frac{x^2}{2!}+\frac{x^4}{4!}-...)\normalsize -b(\large x-\large\frac{x^3}{3!}+\frac{x^5}{5!}-.....)}{x^3}$
$\Rightarrow \large\frac{(1+a-b)x+(b/6-a/2)x^3+(a/24-b/120)x^5+...}{x^3}$
Since given limit is 1(given)
$\therefore 1+a-b=0$ and $b/6-a/2=1$
On solving we get
$a=-\large\frac{5}{2}$ ,$b=-\large\frac{3}{2}$
Hence (c) is the correct answer.
answered Jan 6, 2014 by sreemathi.v
 

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