Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{1-\left[x\right]}{1+x}& ,x\ne 0\\ 1& ,x=0\end{array}$ (where $[.]$ denotes the greatest integer function), then $f\left(x\right)$ is

• A. continuous at $x=\frac{3}{2}$
• B. discontinuous at $x=1$
• C. continuous at $x=\frac{1}{2}$
• D. continuous at $x=0$

Answer 1

Answer: (A), (B), (C)

The correct options are
A continuous at $x=\frac{3}{2}$
B discontinuous at $x=1$
C continuous at $x=\frac{1}{2}$

At $x=\frac{3}{2}$

$\underset{x\to {\frac{3}{2}}^{+}}{lim}f\left(x\right)=\underset{x\to {\frac{3}{2}}^{+}}{lim}\frac{1-\left[x\right]}{1+x}=\frac{1-1}{1+\frac{3}{2}}=0$

$\underset{x\to {\frac{3}{2}}^{-}}{lim}f\left(x\right)=\underset{x\to {\frac{3}{2}}^{-}}{lim}\frac{1-\left[x\right]}{1+x}=\frac{1-1}{1+\frac{3}{2}}=0$

RHL$=$LHL $\therefore$ continuous

Similarly for $x=\frac{1}{2}$

At $x=1$

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

$=\frac{1-1}{1+1},=\frac{1-0}{1+1}$$=0\ne =\frac{1}{2}$

$\therefore$ not continuous

At $x=0$

$f\left(0\right)=1$

$\underset{x\to 0^{+}}{lim}\frac{1-\left[x\right]}{1+x}=\frac{1-0}{1+0}=1$

$\underset{x\to 0^{-}}{lim}\frac{1-\left[x\right]}{1+x}=\frac{1-(-1)}{1+0}=2$

LHL$\ne$RHL $\therefore$discontinuous.