Question

The value of $f\left(0\right)$ , so that the function$f\left(x\right)=\frac{(27-2x{)}^{1/3}-3}{9-3(243+5x{)}^{1/5}};(x\ne 0)$ is continuous at $x=0$ is

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

Answer 1

Answer: (C)

$2$

${lim}_{x\to 0}f\left(x\right)=\frac{{(27-2x)}^{1/3}-3}{9-3{(243+5x)}^{1/5}}$

Applying L'Hospitals rule,
${lim}_{x\to 0}f\left(x\right)=\frac{\left(1/3\right){(27-2x)}^{-2/3}(-2)}{(-3/5){(243+5x)}^{-4/5}\left(5\right)}$

$=\frac{\left(1/3\right){\left(27\right)}^{-2/3}(-2)}{(-3/5){\left(243\right)}^{-4/5}\left(5\right)}=\frac{(-2)\left(1/3\right){\left(27\right)}^{-2/3}}{(-3/5){\left(243\right)}^{-4/5}\left(5\right)}=2$
Hence, option C is the correct answer.