Question

If \[f\left( x \right) = \begin{cases}\frac{\sin\left[ x \right]}{\left[ x \right]}, & \left[ x \right] \neq 0 \\ 0, & \left[ x \right] = 0\end{cases}\]  where  denotes the greatest integer function, then \[\lim_{x \to 0} f\left( x \right)\]  

Options

• 1  

• 0  

• −1     

• does not exist                                    

Answer 1

We have, 

\[\left[ x \right] = \begin{cases}0, & 0 \leq x < 1 \\ - 1, & - 1 \leq x < 0\end{cases}\] 

\[\therefore f\left( x \right) = \begin{cases}\frac{\sin\left( - 1 \right)}{- 1}, & - 1 \leq x < 0 \\ 0, & 0 \leq x < 1\end{cases}\]
\[ \Rightarrow f\left( x \right) = \begin{cases}\sin1, & - 1 \leq x < 0 \\ 0, & 0 \leq x < 1\end{cases}\] 

Now, 

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0} \sin1 = \sin1\]
\[ \lim_{x \to 0^+} f\left( x \right) = \lim_{x \to 0} 0 = 0\]

Clearly, 

\[\lim_{x \to 0^-} f\left( x \right) \neq \lim_{x \to 0^+} f\left( x \right)\]

Thus,

\[\lim_{x \to 0} f\left( x \right)\]does not exist.

Hence, the correct answer is option (d).