Question

Let $f\left(x\right)$ be a function defined as $f\left(x\right)=\{\begin{array}{cc}\frac{x^2-1}{x^2-2|x-1|-1},& x\ne 1\\ \frac{1}{2},& x=1\end{array}$
Discuss the continuity of the function at $x=1$.

Answer 1

$f(1^{\ast }=\underset{x\to 1^{\ast }}{lim}\frac{x^2-1}{x^2-2|x-1|-1}$
$=\underset{x\to 1^{\ast }}{lim}\frac{x^2-1}{x^2-2(x-1)-1}$
$=\underset{x\to 1^{\ast }}{lim}\frac{(x+1)}{(x+1)-2}=\infty$
$f\left(1^{-}\right)=\underset{x\to 1^{-}}{lim}\frac{x^2-1}{x^2-2|x-1|-1}$
$=\underset{x\to 1^{-}}{lim}\frac{x^2-1}{x^2-2(1-x)-1}$
$=\underset{x\to 1^{\ast }}{lim}\frac{(x+1)}{(x+1)+2}=\frac{1}{2}$
Hence, $f\left(x\right)$ is discontinuous at $x=1$.