Question

A function f(x) is defined by $f\left(x\right)=\frac{\left[x^2\right]-1}{x^2-1}$ for $x^2\ne 1$, $f\left(x\right)=f(-1)=0$,where $\left[x\right]$ denotes greatest integer function, then

• A. $\underset{x\to 1^{+}}{lim}f\left(x\right)=0$
• B. $\underset{x\to 1^{-}}{lim}f\left(x\right)=1$
• C. $\underset{x\to 1^{-}}{lim}f\left(x\right)$ doesn't exist
• D. $\underset{x\to 1}{lim}f\left(x\right)$ doesn't exist

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\underset{x\to 1^{+}}{lim}f\left(x\right)=0$
C $\underset{x\to 1^{-}}{lim}f\left(x\right)$ doesn't exist
D $\underset{x\to 1}{lim}f\left(x\right)$ doesn't exist

For $1<x<\sqrt{2},\left[x^2\right]=1$, and for $0<x<1,\left[x^2\right]=0$, therefore
$\underset{x\to 1^{+}}{lim}f\left(x\right)=\frac{\left[x^2\right]-1}{x^2-1}=\underset{x\to 1^{+}}{lim}\frac{1-1}{x^2-1}=0$

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\frac{\left[x^2\right]-1}{x^2-1}=\underset{x\to 1^{-}}{lim}\frac{0-1}{x^2-1}=\underset{x\to 1^{-}}{lim}\frac{-1}{x^2-1}$ which doesn't exist.