Question

Let $f\left(x\right)=\frac{{(27-2x)}^{1/3}-3}{9-3{(243+5x)}^{1/5}},x\ne 0.$ If $f\left(x\right)$ is continuous at $x=0$, then the value of $f\left(0\right)$ is

Answer 1

For continuity of $f\left(x\right)$ at $x=0$, we must have:
$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)$
$=\underset{x\to 0}{lim}\frac{(27-2x{)}^{1/3}-3}{9-3(243+5x{)}^{1/5}}$ $\left[\frac{0}{0}\text{form}\right]$
$=\underset{x\to 0}{lim}\frac{\frac{1}{3}(27-2x{)}^{-2/3}\cdot (-2)}{-\frac{3}{5}(243+5x{)}^{-4/5}\cdot 5}$
$=\frac{2}{9}\underset{x\to 0}{lim}\frac{(243+5x{)}^{4/5}}{(27-2x{)}^{2/3}}$
$=\frac{2}{9}\cdot \frac{(243{)}^{4/5}}{(27{)}^{2/3}}=\frac{2}{9}\cdot \frac{81}{9}=2$