Question

A function f(x) is defined by $f\left(x\right)=\{\begin{array}{cc}\frac{[x^2-1]}{x^2-1}& for{x}^2\ne 1\\ 0& for{x}^2=1\end{array}$ Discuss the contiuuity of f(x) at x=1.

Answer 1

we redefine the function as under.$f\left(x\right)=\{\begin{array}{cc}\frac{-1}{x^2-1},& x<1\\ 0,& x=1\\ \frac{1-1}{x^2-1}=0,& x>1\end{array}$
$R=V=1$
but$L=\stackrel{Lt}{h\to 0}\frac{-1}{{(1-h)}^2-1}$ $=\stackrel{Lt}{h\to 0}\frac{-1}{-2h+h^2}=-\infty$
Since $R\ne L$ therefore $f\left(x\right)$ is not continuous at $x=1.$