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If $f(x)=\left\{\begin{array}{1 1}\large\frac{\sin [x]}{[x]}&[x]\neq 0\\0&[x]=0\end{array}\right.$, where $[x]$ denotes the greatest integer less than or equal to x, then $\lim\limits_{x\to 0}f(x)$ equals


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$f(x)=\left\{\begin{array}{1 1}\large\frac{\sin [x]}{[x]}&if\;x\in(-\infty,0)\cup [1,\infty]\\0&if\;x\in[0,1]\end{array}\right.$
$\lim\limits_{x\to 0^-}f(x)=\lim\limits_{h\to 0}\large\frac{\sin[-h]}{[-h]}$
$\qquad\quad\;\;\;=\lim\limits_{h\to 0}\large\frac{\sin[-1]}{[-1]}$
$\qquad\quad\;\;\;=\sin 1$
And $\lim\limits_{x\to 0^+}f(x)=\lim\limits_{h\to 0}0=0$
$\lim\limits_{x\to 0^-}f(x)\neq\lim\limits_{x\to 0^+}f(x)$ as $\sin 1\neq 0$
$\lim\limits_{x\to 0}f(x)$ does not exist.
Hence (d) is the correct answer.
answered Dec 31, 2013 by sreemathi.v

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