If  $f\left(x\right)=\frac{x^2-bx+25}{x^2-7x+10}$  for $x\ne 5$  is continuous at $x=5,$ then the value of $f\left(5\right)$ is

$f\left(x\right)=\frac{x^2-bx+25}{x^2-7x+10},x\ne 5$

$f\left(x\right)$ is continuous at$x=5,$ only if $\underset{x\to 5}{\lim }\frac{x^2-bx+25}{x^2-7x+10}$  is finite

Now $x^2-7x+10\to 0$ when $x\to 5,$  then we must have $x^2-bx+25\to 0$  for which $b=10$

Hence, $\underset{x\to 5}{\lim }\frac{x^2-10x+25}{x^2-7x+10}=\underset{x\to 5}{\lim }\frac{x-5}{x-2}=0$