If $f(x)=\left\{\begin{array}{c}\frac{x^2-9}{x-3}, x \neq 3 \\ ~5~~, ~~x=3\end{array}\right.$ then $f(x)$:

has removable discontinuity at x = 3

The correct answer is Option (2) - has removable discontinuity at x = 3

$f(5)=5$

so $\lim\limits_{x→3}\frac{x^2-9}{x-3}=\lim\limits_{x→3}\frac{(x-3)(x+3)}{(x-3)}=\lim\limits_{x→3} x+3=6≠f(3)$

⇒ it has removable discontinuity at x = 3

as f(x) can be redefined as 

$f(x)=\left\{\begin{array}{c}\frac{x^2-9}{x-3}, x \neq 3 \\ ~6~~, ~~x=3\end{array}\right.$