Question

Given:$f\left(x\right)=\frac{\log x-\log 3}{x-3}$, for $x\ne 3$ if $f\left(x\right)$ is continuous at $x=3$, find $f\left(3\right)$.

Answer 1

$f\left(x\right)=\frac{\ln x-\ln 3}{x-3}$
f is continuous at $x=3$
${lim}_{x\to 3}\frac{\ln x-\ln 3}{x-3}=f\left(3\right)$
$f\left(3\right)={lim}_{x\to 3}\frac{1}{x}$ (L Hospital rule)
$=\frac{1}{3}$
$f\left(3\right)=\frac{1}{3}$