If lim _{x} {rightarrow 1} frac{(5 x+1)^{1 / | vedclass

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(A) Using $L$'Hopital's rule,we differentiate the numerator and denominator with respect to $x$:$\lim _{x \rightarrow 1} \frac{\frac{d}{dx}((5 x+1)^{1

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) Using $L$'Hopital's rule,we differentiate the numerator and denominator with respect to $x$:$\lim _{x \rightarrow 1} \frac{\frac{d}{dx}((5 x+1)^{1 / 3}-(x+5)^{1 / 3})}{\frac{d}{dx}((2 x+3)^{1 / 2}-(x+4)^{1 / 2})}$$= \lim _{x}$ ${\rightarrow 1} \frac{\frac{1}{3}(5 x+1)^{-2 / 3} \cdot 5 - \frac{1}{3}(x+5)^{-2 / 3}}{\frac{1}{2}(2 x+3)^{-1 / 2} \cdot 2 - \frac{1}{2}(x+4)^{-1 / 2}}$$= \frac{\frac{5}{3}(6)^{-2 / 3} - \frac{1}{3}(6)^{-2 / 3}}{(5)^{-1 / 2} - \frac{1}{2}(5)^{-1 / 2}}$$= \frac{\frac{4}{3} \cdot 6^{-2 / 3}}{\frac{1}{2} \cdot 5^{-1 / 2}} = \frac{8}{3} \cdot 6^{-2 / 3} \cdot \sqrt{5} = \frac{8 \sqrt{5}}{3 \cdot 6^{2 / 3}}$Since $6^{2 / 3} = (2 \cdot 3)^{2 / 3} = 2^{2 / 3} \cdot 3^{2 / 3}$,the expression becomes $\frac{8 \sqrt{5}}{3^{1/3} \cdot 2^{2/3}}$.Comparing with $\frac{m \sqrt{5}}{n(2 n)^{2 / 3}}$,we have $m=8$ and $n=3$.Thus,$8m + 12n = 8(8) + 12(3) = 64 + 36 = 100$.

If $\lim _{x}$ ${\rightarrow 1} \frac{(5 x+1)^{1 / 3}-(x+5)^{1 / 3}}{(2 x+3)^{1 / 2}-(x+4)^{1 / 2}}=\frac{m \sqrt{5}}{n(2 n)^{2 / 3}}$,where $\operatorname{gcd}(m, n)=1$,then $8 m+12 n$ is equal to.

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