lim limits_{x} {rightarrow a} frac{(a+2x)^{1/ | vedclass

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(A) Let $L = \lim_{x \rightarrow a} \frac{(a+2x)^{1/3}-(3x)^{1/3}}{(3a+x)^{1/3}-(4x)^{1/3}}$.Using $L$'$H$ôpital's rule,we differentiate the numerator

Vedclass · Accepted Answer

$\left(\frac{2}{3}\right)\left(\frac{2}{9}\right)^{1/3}$

Vedclass · Answer

(A) Let $L = \lim_{x \rightarrow a} \frac{(a+2x)^{1/3}-(3x)^{1/3}}{(3a+x)^{1/3}-(4x)^{1/3}}$.Using $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$:Numerator derivative: $\frac{d}{dx} ((a+2x)^{1/3} - (3x)^{1/3}) = \frac{1}{3}(a+2x)^{-2/3}(2) - \frac{1}{3}(3x)^{-2/3}(3) = \frac{2}{3}(a+2x)^{-2/3} - (3x)^{-2/3}$.Denominator derivative: $\frac{d}{dx} ((3a+x)^{1/3} - (4x)^{1/3}) = \frac{1}{3}(3a+x)^{-2/3} - \frac{1}{3}(4x)^{-2/3}(4) = \frac{1}{3}(3a+x)^{-2/3} - \frac{4}{3}(4x)^{-2/3}$.Evaluating at $x = a$:Numerator: $\frac{2}{3}(3a)^{-2/3} - (3a)^{-2/3} = (3a)^{-2/3} (\frac{2}{3} - 1) = -\frac{1}{3}(3a)^{-2/3}$.Denominator: $\frac{1}{3}(4a)^{-2/3} - \frac{4}{3}(4a)^{-2/3} = (4a)^{-2/3} (\frac{1}{3} - \frac{4}{3}) = -(4a)^{-2/3}$.Thus,$L = \frac{-\frac{1}{3}(3a)^{-2/3}}{-(4a)^{-2/3}} = \frac{1}{3} \left(\frac{4a}{3a}\right)^{2/3} = \frac{1}{3} \left(\frac{4}{3}\right)^{2/3} = \frac{1}{3} \frac{4^{2/3}}{3^{2/3}} = \frac{4^{2/3}}{3^{5/3}} = \frac{2^{4/3}}{3^{5/3}} = \frac{2 \cdot 2^{1/3}}{3 \cdot 3^{2/3}} = \frac{2}{3} \left(\frac{2}{9}\right)^{1/3}$.

$\lim \limits_{x}$ ${\rightarrow a} \frac{(a+2x)^{1/3}-(3x)^{1/3}}{(3a+x)^{1/3}-(4x)^{1/3}} \text{ for } a \neq 0 \text{ is equal to}$

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