The function $f(x)=\left\{\begin{matrix} \frac{x^2+2x-3}{x-1} & ,x≠1\\2 & ,x=1 \end{matrix}\right.$ is :

discontinuous at x = 1

The correct answer is Option (2) → discontinuous at x = 1

$f(1)=2$

$\lim\limits_{x→1}f(x)=\lim\limits_{x→1}\frac{x^2+2x-3}{x-1}$

$=\lim\limits_{x→1}\frac{x^2+3x-x-3}{x-1}$

$=\lim\limits_{x→1}\frac{(x+3)(x-1)}{x-1}$

$=\lim\limits_{x→1}(x+3)=4≠f(1)$

discontinuous at $x=1$