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If $f(x)=\sqrt{\large\frac{x-\sin x}{x+\cos^2x}}$ then $\lim\limits_{x\to \infty}f(x)$ is

$(a)\;0\qquad(b)\;\infty\qquad(c)\;1\qquad(d)\;None\;of\;these$

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$f(x)=\sqrt{\large\frac{x-\sin x}{x+\cos^2x}}$
$\Rightarrow \lim\limits_{x\to \infty}f(x)=\lim\limits_{x\to \infty}\sqrt{\large\frac{1-\large\frac{\sin x}{x}}{1+\Large \frac{\cos^2x}{x}}}$
$\Rightarrow \sqrt{\large\frac{1-0}{1+0}}$
$\Rightarrow 1$
Hence (c) is the correct option.
answered Dec 20, 2013 by sreemathi.v
 

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