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If $f(x)=\left\{\begin{array}{1 1}\large\frac{\sin[x]}{[x]}&[x]\neq 0\\0&[x]=0\end{array}\right.$ then

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

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$\lim\limits_{x\to 0^-}f(x)=\lim\limits_{x\to 0^-}\large\frac{\sin[x]}{[x]}$
$\Rightarrow \large\frac{\sin(-1)}{-1}$
$\Rightarrow -\sin 1$
$\lim\limits_{x\to 0^+}f(x)=\lim\limits_{x\to 0^+}\large\frac{\sin[x]}{[x]}$
$\Rightarrow \large\frac{\sin(1)}{1}$
$\therefore \lim\limits_{x\to 0^+}f(x)$ does not exist.
Hence (d) is the correct answer.
answered Dec 24, 2013 by sreemathi.v
 

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