If f(x) = l [x] [x] , when , , [x] 0 , , , , , , , , ,0, when , ,… - Vidyadip

MCQ

If $f(x) = \left\{ \begin{array}{l}\frac{{\sin [x]}}{{[x]}},{\rm{ when\,\, }}[x] \ne 0\\\,\,\,\,\,\,\,\,\,0,{\rm{ when \,\,}}[x] = 0\end{array} \right.$ where $[x]$ is greatest integer function, then $\mathop {\lim }\limits_{x \to 0} f(x) = $

A
$-1$
B
$1$
C
$0$
✓
None of these

✓

Answer

Correct option: D.

None of these

In closed interval of $x = 0$ at right hand side $[x] = 0$ and at left hand side $[x] = - 1.$
Also $[0]=0.$ Therefore function is defined as $f(x) = \left\{ \begin{array}{l}\frac{{\sin \,[x]}}{{[x]}}\,\,( - 1 \le x < 0)\\\;\;\;\;\;0\;\;(0 \le x < 1)\end{array} \right.$
$\therefore$ Left hand limit $ = \mathop {\lim }\limits_{x \to 0 - } f(x) = \mathop {\lim }\limits_{x \to 0 - } \,\frac{{\sin \,[x]}}{{[x]}}$
$ = \frac{{\sin \,( - 1)}}{{ - 1}} = \sin {1^c}$
Right hand limit $= 0$.
Hence limit doesn't exist.

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