The value of f(0),so that the function f(x) = | vedclass

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(C) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.Given $f(x) = \frac{(27 - 2x)^{1/3} - 3}{9 - 3(243 +

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$2$

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(C) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.Given $f(x) = \frac{(27 - 2x)^{1/3} - 3}{9 - 3(243 + 5x)^{1/5}}$.As $x 	o 0$,the expression takes the indeterminate form $\frac{0}{0}$.Using $L$'$H$ôpital's Rule,we differentiate the numerator and denominator with respect to $x$:Numerator derivative: $\frac{d}{dx} [(27 - 2x)^{1/3} - 3] = \frac{1}{3}(27 - 2x)^{-2/3} \cdot (-2) = -\frac{2}{3}(27 - 2x)^{-2/3}$.Denominator derivative: $\frac{d}{dx} [9 - 3(243 + 5x)^{1/5}] = -3 \cdot \frac{1}{5}(243 + 5x)^{-4/5} \cdot 5 = -3(243 + 5x)^{-4/5}$.Now,taking the limit as $x 	o 0$:$\lim_{x 	o 0} f(x) = \frac{-\frac{2}{3}(27)^{-2/3}}{-3(243)^{-4/5}} = \frac{\frac{2}{3}(3^3)^{-2/3}}{3(3^5)^{-4/5}} = \frac{\frac{2}{3}(3^{-2})}{3(3^{-4})} = \frac{\frac{2}{3} \cdot \frac{1}{9}}{3 \cdot \frac{1}{81}} = \frac{2/27}{3/81} = \frac{2/27}{1/27} = 2$.Thus,$f(0) = 2$.

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

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