Question

If $f\left(x\right)=\left\{\begin{array}{ll}\frac{\sin \left[x\right]}{\left[x\right]},& \left[x\right]\ne 0\\ 0,& \left[x\right]=0\end{array}\right.$, where [.] denotes the greatest integer function, then $\underset{x\to 0}{\lim }f\left(x\right)$ is equal to

(a) 1 (b) 0 (c) −1 (d) does not exist

Answer 1

We have,

$\left[x\right]=\left\{\begin{array}{ll}0,& 0\le x<1\\ -1,& -1\le x<0\end{array}\right.$

$$\begin{gathered} \therefore f\left(x\right)=\left\{\begin{array}{ll}\frac{\sin \left(-1\right)}{-1},& -1\le x<0\\ 0,& 0\le x<1\end{array}\right. \\ \Rightarrow f\left(x\right)=\left\{\begin{array}{ll}\sin 1,& -1\le x<0\\ 0,& 0\le x<1\end{array}\right. \end{gathered}$$

Now,

$$\begin{gathered} \underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\sin 1=\sin 1 \\ \underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }0=0 \end{gathered}$$

Clearly, $\underset{x\to 0^{-}}{\lim }f\left(x\right)\ne \underset{x\to 0^{+}}{\lim }f\left(x\right)$

Thus, $\underset{x\to 0}{\lim }f\left(x\right)$ does not exist.

Hence, the correct answer is option (d).