Question

Let $f\left(x\right)=\left[x\right]\sin \left(\frac{\pi }{[x+1]}\right),$ where $[.]$ denotes the greatest integer function. Then

• A. domain of $f$ is $R-\{-1\}$
• B. $\underset{x\to 0^{+}}{lim}f\left(x\right)=0$
• C. $f$ is continuous on $[0,1)$
• D. $\underset{x\to 1^{+}}{lim}f\left(x\right)=1$

Answer 1

Answer: (B), (C), (D)

$\underset{x\to 1^{+}}{lim}f\left(x\right)=1$

$[x+1]=0$
$\Rightarrow 0\le x+1<1$
$\Rightarrow -1\le x<0$
$\therefore$ Domain of $f$ is $R-[-1,0)$

$\underset{x\to 0^{+}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(0+h)$
$=\underset{h\to 0}{lim}[0+h]\sin \left(\frac{\pi }{[0+h]+1}\right)$
$=\underset{h\to 0}{lim}(0\cdot \sin \pi )=0$

For $x\in [0,1),[x+1]=1$
$\therefore f\left(x\right)=0\cdot \sin \pi =0$ for all $x\in [0,1)$
$\therefore f$ is continuous on $[0,1)$

$\underset{x\to 1^{+}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(1+h)$
$=\underset{h\to 0}{lim}[1+h]\sin \left(\frac{\pi }{[1+h]+1}\right)$
$=1\times \sin \frac{\pi }{2}$
$=1$