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Q.
If $f \left(x\right)=\frac{x^{2}-bx+25}{x^{2}-7x+10}$ for $x \ne5$ is continuous at x = 5, then the value of $f (5)$ is
Continuity and Differentiability
Solution:
$f (x)=\frac{x^{2}-bx+25}{x^{2} -7x+10}$,
$x \ne5$
$f (x)$ is continuous at x = 5 only if $\displaystyle\lim_{x\to5}\frac{x^{2}-bx+25}{x^{2}-7x+10}$ 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, $\displaystyle\lim_{x\to5}\frac{x^{2}-10x+25}{x^{2}-7x+10}$
$=\displaystyle\lim_{x\to5}\frac{x-5}{x-2}=0$