$f(5)=\lim\limits_{x\to 5}f(x)=\lim\limits_{x\to 5}\large\frac{x^2-10x+25}{x^2-7x+10}$
$\Rightarrow \lim\limits_{x\to 5}\large\frac{(x-5)^2}{(x-2)(x-5)}$
$\Rightarrow \lim\limits_{x\to 5}\large\frac{(x-5)}{(x-2)}$
$\Rightarrow \large\frac{5-5}{5-2}$
$\Rightarrow 0$
Hence (a) is the correct answer.