The function $f(x)=\left\{\begin{array}{cl}\frac{x^2+2 x-3}{x-1} & , \text { if } x \neq 1 \\ 0 & , \text { if } x=1\end{array}\right.$ is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) - discontinuous at x = 1
$f(1)=0$
$\lim\limits_{x→1}\frac{x^2+2x-3}{x-1}=\lim\limits_{x→1}\frac{(x-1)(x+3)}{(x-1)}$
$\lim\limits_{x→1}(x+3)=4$
$4≠0$ at x = 1
f(x) is discontinuous