If \[f\left( x \right) = \left\{ \begin{array}{l} \frac{{{{\left[ x \right]}^2} + \sin \left[ x \right]}}{{\left[ x \right]}}\,{\rm{ for }}\left[ x \right] \ne 0\\ \,\,\,\,\,\,\,\,\,0{\rm{ }}\,\,\,\,\,\,\,\,\,\,{\rm{ for }}\left[ x \right] = 0 \end{array} \right.,{\rm{where }}\left[ x \right]\]        denotes the greatest integer less than or equal to $$x,$$ then $$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$$   equals :

Solution :
$$\eqalign{ & {\text{As }}x \to 0 - \left( {{\text{i}}{\text{.e}}{\text{., approaches 0 from the left}}} \right),\,\left[ x \right] = - 1 \cr & \therefore \,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{1 + \sin \left( { - 1} \right)}}{{ - 1}} = - 1 + \sin \,1 \cr & {\text{Whereas, if }}x \to {0^ + }{\text{ we get }}\left[ x \right] = 0 \cr & \therefore \,f\left( x \right) = 0\, \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 0 \cr & {\text{Thus, }}\mathop {\lim }\limits_{x \to 0} f\left( x \right){\text{ does not exist}}{\text{.}} \cr} $$