Question

Find the value of $f\left(0\right)$ so that the function $f\left(x\right)=\frac{96[\log (1+\frac{x}{12})-\log (1-\frac{x}{8})]}{x},x\ne 0$ is continuous on $[0,8]$

Answer 1

$f\left(0\right)=\underset{x\to 0}{lim}\frac{96\left[\log \right(1+\frac{x}{12})-\log (1-\frac{x}{8}\left)\right]}{x}$
$=96\underset{x\to 0}{lim}\cdot \frac{\left[\log \right(1+\frac{x}{12})}{12\left(\frac{x}{12}\right)}+\frac{\log (1-\frac{x}{8})]}{8(-\frac{x}{8})}$
$=96(\frac{1}{12}+\frac{1}{8})=20$