A function $f\left(x\right)$ is defined by, $f\left(x\right)=\left\{\begin{array}{r}\frac{\left[x^2\right]-1}{x^2-1},\text{ for }{x}^2\ne 1\\ 0,\text{ for }{x}^2=1\end{array}\right.$ 

Then, the continuity of $f\left(x\right)$ at $x=1$ is 

The given function is $f\left(x\right)=\left\{\begin{array}{r}\frac{\left[x^2\right]-1}{x^2-1},\text{ for }{x}^2\ne 1\\ 0,\text{ for }{x}^2=1\end{array}\right.$

                                 $f\left(x\right)=\left\{\begin{array}{l}\frac{-1}{x^2-1} ,for x^2\ne 1\\ 0 , for x^2=1\end{array}\right.$

Right Hand Limit at $x=1$ is given by

$\underset{x\to 1^{+}}{\lim }f\left(x\right)=0$

Also, Left Hand Limit at $x^2=1$ is given by

$$\begin{gathered} \underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f(1-h) \\ =\underset{h\to 0}{\lim }\frac{-1}{(1-h{)}^2-1} \\ =-\infty \end{gathered}$$

$\underset{x\to 1}{\lim }f\left(x\right)$ doesn't exist because Right Hand limit is not equal to Left Hand limit.

Hence, $f\left(x\right)$ is not continuous at $x=1$.

Therefore, the correct answer is option $\left(1\right)$.