Question

The value of $\underset{x\to 0}{lim}\left[\frac{\sin [x-3]}{[x-3]}\right]$ is
(where $[.]$ represents the greatest integer function)

• A. $0$
• B. $1$
• C. $\sin 1$
• D. does not exist

Answer 1

The correct option is D does not exist

$L.H.L.=\underset{x\to 0^{-}}{lim}\left[\frac{\sin [x-3]}{[x-3]}\right]$
$=\underset{h\to 0}{lim}\left[\frac{\sin \left(\right[0-h-3\left]\right)}{[0-h-3]}\right]$
$=\underset{h\to 0}{lim}\left[\frac{\sin \left(\right[-h-3\left]\right)}{[-h-3]}\right]$
$=\left[\frac{\sin (-4)}{-4}\right]$
$=\left[\frac{\sin \left(4\right)}{4}\right]=-1$
$(\because \sin 4<0\text{as}\pi <4<\frac{3\pi }{2})$

$R.H.L.=\underset{x\to 0^{+}}{lim}\left[\frac{\sin \left(\right[x-3\left]\right)}{[x-3]}\right]$
$=\underset{h\to 0}{lim}\left[\frac{\sin \left(\right[0+h-3\left]\right)}{[0+h-3]}\right]$
$=\left[\frac{\sin (-3)}{-3}\right]$
$=\left[\frac{\sin \left(3\right)}{3}\right]=0$
$(\because \sin 3>0\text{as}\frac{\pi }{2}<3<\pi )$

$\Rightarrow L.H.L.\ne R.H.L.$
Hence, limit does not exist.