Question

f (x) be a function of x defined as $f\left(x\right)=\frac{x^2-1}{x^2-2|x-1|-1},x\ne 1,=1/2,x=1$ Is the function continuous at x=1?

Answer 1

LHL=$f(1-h)={lim}_{h\to 0}\frac{{(1-h)}^2-1}{{(1-h)}^2-2|(1-h)-1|-1}={lim}_{h\to 0}\frac{h(h-2)}{h(h-4)}=\frac{1}{2}$

V=$f\left(1\right)=1/2$

RHL=$f(1+h)=\underset{h\to 0}{lim}\frac{{(1-h)}^2-1}{{(1-h)}^2-2\left|(1+h-1)\right|-1}$ $=\underset{h\to 0}{lim}\frac{h^2+2h}{h^2}=\underset{h\to 0}{lim}(1+\frac{2}{h})=\infty$

Since $RHL\ne LHL=V$
$\therefore$ function is discontinuous at $x=1.$