Question
If $$\eqalign{
& f\left( x \right) = \frac{{\sin \left[ x \right]}}{{\left[ x \right]}},\,\,\left[ x \right] \ne 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ x \right] = 0 \cr} $$
Where \[\left[ x \right]\] denotes the greatest integer less than or equal to $$x.$$ then $$\mathop {\lim }\limits_{x\, \to \,0} f\left( x \right)$$ equals
A.
$$1$$
B.
$$0$$
C.
$$ - 1$$
D.
none of these
Answer :
none of these
Solution :
The given function can be restated as
\[f\left( x \right) = \left\{ \begin{array}{l}
\frac{{\sin \left[ x \right]}}{{\left[ x \right]}},\,\,\,\,{\rm{if}}\,x \in \left( { - \,\infty ,0} \right) \cup \left[ {1,\,\infty } \right]\\
0\,\,\,\,\,,\,\,\,\,\,\,\,{\rm{if}}\,x \in \left[ {0,\,1} \right)
\end{array} \right.\]
$$\eqalign{
& \therefore \mathop {\lim }\limits_{x\, \to \,{0^ - }} f\left( x \right) = \mathop {\lim }\limits_{h\,\, \to \,0} \frac{{\sin \left[ { - h} \right]}}{{\left[ { - h} \right]}} \cr
& = \mathop {\lim }\limits_{x\, \to \,{0^ - }} \frac{{\sin \left( { - 1} \right)}}{{\left( { - 1} \right)}} = \sin \,1 \cr
& {\text{And}}\mathop {\lim }\limits_{x\, \to \,{0^ + }} f\left( x \right) = \mathop {\lim }\limits_{h\, \to \,0} 0 = 0 \cr
& \because \mathop {\lim }\limits_{x\, \to \,{0^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x\, \to \,{0^ + }} f\left( x \right)\,\,\,{\text{as}}\,\,\sin \,1 \ne 0 \cr
& \therefore \mathop {\lim }\limits_{x\, \to \,0} f\left( x \right)\,\,{\text{does not exist}}{\text{.}} \cr} $$