If $f(x)=\frac{x^2-10 x+25}{x^2-7 x+10}$ and $f$ is continuous at $x=5$, then $f(5)$ is equal to

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Aicharya · Accepted Answer

$\begin{aligned} f(5) & =\lim _{x ightarrow 5} f(x)=\lim _{x ightarrow 5} \frac{x^2-10 x+25}{x^2-7 x+10} \ & =\lim _{x ightarrow 5} \frac{(x-5)^2}{(x-5)(x-2)}=\lim _{x ightarrow 5} \frac{x-5}{x-2} \ & =\frac{5-5}{5-2}=0\end{aligned}$

If $f(x)=\frac{x^2-10 x+25}{x^2-7 x+10}$ and $f$ is continuous at $x=5$, then $f(5)$ is equal to

Solution