The value of f(0), so that the function


The value of $f(0)$, so that the function

$f(x)=\frac{(27-2 x)^{1 / 3}-3}{9-3(243+5 x)^{1 / 5}}(x \neq 0)$ is continuous, is given by

(a) $\frac{2}{3}$

(b) 6

(c) 2

(d) 4


(c) 2

For f(x) to be continuous at x = 0, we must have

$\lim _{x \rightarrow 0} f(x)=f(0)$

$\Rightarrow f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{(27-2 x)^{\frac{1}{3}-3}}{9-3(243+5 x)^{\frac{1}{5}}}$

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

$=\frac{2}{15} \times \frac{\frac{1}{3} \times 27 \frac{\frac{-2}{3}}{\frac{1}{5} \times 243^{\frac{-4}{5}}}}{\frac{1}{5}}$

$=\frac{2}{15} \times \frac{\frac{1}{3} \times \frac{1}{27^{\frac{2}{3}}}}{\frac{1}{5} \times \frac{1}{243^{\frac{4}{5}}}}$


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