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The value of $\lim\limits_{x\to 0}\int\limits_{0^{x^2}}\large\frac{\cos t^2dt}{x\sin x}$ is

$(a)\;\large\frac{3}{2}$$\qquad(b)\;1\qquad(c)\;-1\qquad(d)\;None\;of\;these$

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$\lim\limits_{x\to 0}\int_0^{x^2}\large\frac{\cos t^2dt}{x\sin x}\qquad\big(\large\frac{0}{0}\big)$
By L Hospitals rule
$\Rightarrow \lim\limits_{x\to 0}\large\frac{2x\cos x^4}{x\cos x+\sin x}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{2\cos x^4-\sin x^4}{2\cos x-x\sin x}$
$\Rightarrow \large\frac{2-0}{2-0}$
$\Rightarrow 1$
Hence (b) is the correct answer.
answered Dec 30, 2013 by sreemathi.v
 

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