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Q.
If $f(x)=\frac{x^{2}-b x+25}{x^{2}-7 x+10}$ for $x \neq 5$ is continuous at $x=5$, then the value of $f(5)$ is
Continuity and Differentiability
Solution:
$f(x)=\frac{x^{2}-b x+25}{x^{2}-7 x+10}, x \neq 5$
$f(x)$ is continuous at $x=5$, only if $\displaystyle\lim _{x \rightarrow 5} \frac{x^{2}-b x+25}{x^{2}-7 x+10}$ is finite.
Now $x^{2}-7 x+10 \rightarrow 0$ when $x \rightarrow 5$,
then we must have $x^{2}-b x$ $+25 \rightarrow 0$ for which $b=10$
Hence, $\displaystyle\lim _{x \rightarrow 5} \frac{x^{2}-10 x+25}{x^{2}-7 x+10}$
$=\displaystyle\lim _{x \rightarrow 5} \frac{x-5}{x-2}=0$