Question

If $f\left(x\right)=\frac{x^2-bx+36}{x^2-9x+18}$ for $x\ne 6$ is continuous at $x=6$, then value of $f\left(6\right)$ is

Answer 1

$f\left(x\right)$ is continuous at $x=6$ only if
$\underset{x\to 6}{lim}\frac{x^2-bx+36}{x^2-9x+18}$ is finite.
As $x^2-9x+18\to 0$ as $x\to 6$
Then we must have $x^2-bx+36\to 0$ as $x\to 0$

$\Rightarrow b=12$
hence, $\underset{x\to 6}{lim}\frac{x^2-bx+36}{x^2-9x+18}=\underset{x\to 6}{lim}\frac{(x-6)(x-6)}{(x-6)(x-3)}$
$=\underset{x\to 6}{lim}\frac{x-6}{x-3}=0$