Question

A Funtion f is defined as $f\left(x\right)=\frac{x^2-4x+3}{x^2-1}$ for $x\ne 1,=2$ for $x=1$ Is the function continuous at $x=1.$?

• A. True
• B. False

Answer 1

The correct option is B False

We have given ,$f\left(1\right)=2$
$f\left(x\right)=\frac{(x-1)(x-3)}{(x-1)(x+1)}=\frac{x-3}{x+1},x\ne 1$
$R.H.L.=\underset{x\to 0^{+}}{lim}\frac{1+h-3}{1+h+1}=\frac{-2}{2}=-1$
$L.H.L.=\underset{x\to 0^{-}}{lim}\frac{1-h-3}{1-h+1}=\frac{-2}{2}=-1$
Since
$f\left(1\right)\ne f\left(1^{+}\right)=f\left(1^{-}\right),$ the function is discontinuous at $x=1$.