Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{x^2-4x+3}{x^2-1},& forx\ne 1\\ 2,& forx=1\end{array}$ , then

• A. ${lim}_{x\to 1+}f\left(x\right)=2$
• B. ${lim}_{x\to 1-}f\left(x\right)=3$
• C. f(x) is discontinuous at x = 1
• D. None of these

Answer 1

The correct option is C f(x) is discontinuous at x = 1

$f\left(x\right)=\left\{\frac{x^2-4x+3}{x^2-1}\right\}$ for $x\ne 1$
=2, for x = 1
$f\left(1\right)=2,f(1+)={lim}_{x\to 1+}\frac{x^2-4x+3}{x^2-1}={lim}_{x\to 1+}\frac{(x-3)}{(x+1)}=-1$
$f(1-)={lim}_{x\to 1-}\frac{x^2-4x+3}{x^2-1}=-1\Rightarrow f\left(1\right)\ne f(1-)$
Hence the function is discontinuous at x = 1