lim _{x rightarrow 3} frac{(84-x)^{frac{1}{4} | vedclass

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Question

(A) Let $f(x) = \frac{(84-x)^{\frac{1}{4}}-3}{x-3}$.As $x \rightarrow 3$,the numerator becomes $(84-3)^{\frac{1}{4}}-3 = 81^{\frac{1}{4}}-3 = 3-3 = 0$

Vedclass · Accepted Answer

$\frac{-1}{108}$

Vedclass · Answer

(A) Let $f(x) = \frac{(84-x)^{\frac{1}{4}}-3}{x-3}$.As $x ightarrow 3$,the numerator becomes $(84-3)^{\frac{1}{4}}-3 = 81^{\frac{1}{4}}-3 = 3-3 = 0$ and the denominator becomes $3-3 = 0$.This is a $\frac{0}{0}$ indeterminate form.Using $L$'Hopital's Rule,we differentiate the numerator and denominator with respect to $x$:$\frac{d}{dx} ((84-x)^{\frac{1}{4}}-3) = \frac{1}{4}(84-x)^{-\frac{3}{4}} 	imes (-1) = -\frac{1}{4}(84-x)^{-\frac{3}{4}}$.$\frac{d}{dx} (x-3) = 1$.Now,evaluate the limit as $x ightarrow 3$:$\lim _{x}$ ${ightarrow 3} \frac{-\frac{1}{4}(84-x)^{-\frac{3}{4}}}{1} = -\frac{1}{4}(84-3)^{-\frac{3}{4}} = -\frac{1}{4}(81)^{-\frac{3}{4}}$.Since $81 = 3^4$,we have $81^{-\frac{3}{4}} = (3^4)^{-\frac{3}{4}} = 3^{-3} = \frac{1}{27}$.Therefore,the limit is $-\frac{1}{4} 	imes \frac{1}{27} = -\frac{1}{108}$.

$\lim _{x \rightarrow 3} \frac{(84-x)^{\frac{1}{4}}-3}{x-3}$ is

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