Question

The value of f (0), so that the function
$f\left(x\right)=\frac{{\left(27-2x\right)}^{1/3}-3}{9-3{\left(243+5x\right)}^{1/5}}\left(x\ne 0\right)$ is continuous, is given by
(a) $\frac{2}{3}$
(b) 6
(c) 2
(d) 4

Answer 1

(c) 2

For f(x) to be continuous at x = 0, we must have
$$\begin{gathered} \underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right) \\ \Rightarrow f\left(0\right)=\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\frac{{\left(27-2x\right)}^{{\displaystyle \frac{1}{3}}}-3}{9-3{\left(243+5x\right)}^{{\displaystyle \frac{1}{5}}}} \\ \Rightarrow f\left(0\right)=\underset{x\to 0}{\lim }\frac{{\left(27-2x\right)}^{{\displaystyle \frac{1}{3}}}-{27}^{{\displaystyle \frac{1}{3}}}}{3\left({243}^{{\displaystyle \frac{1}{5}}}-{\left(243+5x\right)}^{{\displaystyle \frac{1}{5}}}\right)} \\ =\frac{1}{3}\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{{\left(27-2x\right)}^{{\displaystyle \frac{1}{3}}}-{27}^{{\displaystyle \frac{1}{3}}}}{x}}}{{\displaystyle \frac{\left({243}^{{\displaystyle \frac{1}{5}}}-{\left(243+5x\right)}^{{\displaystyle \frac{1}{5}}}\right)}{x}}} \\ =\frac{-1}{3}\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{{\left(27-2x\right)}^{{\displaystyle \frac{1}{3}}}-{27}^{{\displaystyle \frac{1}{3}}}}{x}}}{{\displaystyle \frac{\left({\left(243+5x\right)}^{{\displaystyle \frac{1}{5}}}-{243}^{\frac{1}{5}}\right)}{x}}} \\ =\frac{2}{15}\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{{\left(27-2x\right)}^{{\displaystyle \frac{1}{3}}}-{27}^{{\displaystyle \frac{1}{3}}}}{-2x}}}{{\displaystyle \frac{\left({\left(243+5x\right)}^{{\displaystyle \frac{1}{5}}}-{243}^{\frac{1}{5}}\right)}{5x}}} \\ =\frac{2}{15}\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{{\left(27-2x\right)}^{{\displaystyle \frac{1}{3}}}-{27}^{{\displaystyle \frac{1}{3}}}}{27-2x-27}}}{{\displaystyle \frac{\left({\left(243+5x\right)}^{{\displaystyle \frac{1}{5}}}-{243}^{\frac{1}{5}}\right)}{243+5x-243}}} \\ =\frac{2}{15}\times \frac{{\displaystyle \frac{1}{3}}\times {27}^{{\displaystyle \frac{-2}{3}}}}{{\displaystyle \frac{1}{5}}\times {243}^{{\displaystyle \frac{-4}{5}}}} \\ =\frac{2}{15}\times \frac{{\displaystyle \frac{1}{3}\times \frac{1}{{27}^{\frac{2}{3}}}}}{{\displaystyle \frac{1}{5}\times \frac{1}{{243}^{\frac{4}{5}}}}} \\ =2 \end{gathered}$$