Question

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 $f(5)$ is

• A. $0$
• B. $5$
• C. $10$
• D. $25$

Answer 1

Answer: (A)

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