Question

The value of $f\left(0\right)$, so that the function f defined as

$f\left(x\right)=\frac{(27-2x{)}^{\frac{1}{3}}-3}{9-3(243+5x{)}^{\frac{1}{5}}},x\ne 0$

is continuous at $x=0$, is given by

• A. $\frac{2}{3}$
• B. 6
• C. 4
• D. 2

Answer 1

The correct option is D 2

We have

$f\left(0\right)={lim}_{x\to 0}\frac{(27-2x{)}^{\frac{1}{3}}-{27}^{\frac{1}{3}}}{{243}^{1/5}-(243+5x{)}^{\frac{1}{5}}}$

$={lim}_{x\to 0}\frac{\frac{(27-2x{)}^{\frac{1}{3}}-{27}^{\frac{1}{3}}}{27-2x-27}\cdot (27-2x-27)}{\frac{(243{)}^{1/5}-(243+5x{)}^{1/5}}{243-243-5x}\cdot (243-243-5x)}$

$=\frac{\frac{1}{3}(27{)}^{-2/3}}{3\cdot \frac{1}{5}(243{)}^{-4/5}}\cdot \frac{-2}{-5}=2$