Question

If $f\left(x\right)=\frac{\sin \left[x\right]}{\left[x\right]},\left[x\right]\ne 0$
= 0 , [ x ] = 0
Where [x] denotes the greatest integer less than or equal to $x$. then $\underset{x\to 0}{\lim }f(x)$ equals -

• A. 1
• B. 0
• C. -1
• D. none of these

Answer 1

Answer: (D)

The given function can be restated as $f(x)=\{\begin{array}{ll}\frac{\sin [x]}{[x]},& \text{if }x\in (-\infty ,0)\cup [1,\infty ]\\ 0& \text{if }x\in [0,1]\end{array}$
$\therefore \underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }\frac{\sin \left[-h\right]}{\left[-h\right]}$
$=\underset{x\to 0^{-}}{\lim }\frac{\sin \left(-1\right)}{\left(-1\right)}=\sin 1$
And $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }0=0$
$\because \underset{x\to 0^{-}}{\lim }f\left(x\right)\ne \underset{x\to 0^{+}}{\lim }f\left(x\right)$ as $\sin 1\ne 0$
$\therefore \underset{x\to 0}{\lim }f\left(x\right)$ does not exist.