Question

$\underset{x\to 0}{\lim }\left[\frac{sin\left[x-3\right]}{\left[x-3\right]}\right]$, where [ . ] denotes greatest integer function is

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

Answer 1

Answer: (C)

$\underset{x\to 0}{\lim }\left[\frac{sin\left[x-3\right]}{\left[x-3\right]}\right]$
For $x\to 0^{+},\left[x-3\right]=-3$
$\therefore \frac{sin\left[x-3\right]}{\left[x-3\right]}=\frac{sin\left(-3\right)}{-3}=\frac{sin3}{-3}\in \left(0,1\right)$
$\therefore \underset{x\to 0^{+}}{\lim }\frac{sin\left[x-3\right]}{\left[x-3\right]}=0$
For $x\to 0^{-},\left[x-3\right]=-4$
$\therefore \frac{sin\left[x-3\right]}{\left[x-3\right]}=\frac{sin4}{4}$ lies in $\left(-1,0\right)$
$\therefore \underset{x\to 0^{-}}{\lim }\frac{sin\left[x-3\right]}{\left[x-3\right]}=-1\therefore$ Limit does not exist.