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 

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

Now, $x^2-7x+10\to 0 \text{when }x\to 5.$ Then we must

have $x^2-bx+25\to 0$ for which b = 10.

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