If x 1lim(5x+1)^1/3-(x+5)^1/3(2x+3)^1/2-(x+4)^1/2=m√5n(2n)^2/3, where gcd(m,n)=1, then 8m+…

Question

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

Answer: 100

Accepted Answer

$I=\underset{x	o 1}{lim}\frac{(5x+1{)}^{1/3}-(+5{)}^{1/}}{(2x+3{)}^{1/2}-(x+4{)}^{1/2}}$

From: $\frac{0}{0}$, using $L-H$ rule

$\begin{gathered} I=\underset{x	o 1}{lim}\frac{\frac{1}{3}	imes 5(5x+1{)}^{-2/3}--(+5)}{\frac{1}{2}	imes 2(2x+3{)}^{-1/2}-\frac{1}{2}(x+4{)}^{-1/2}} \ =\frac{(\frac{5}{3}-\frac{1}{3}|\, {6}^{-2/3}}{\frac{1}{2}{5}^{-1/2}}=\frac{8}{3}	imes \frac{{5}^{1/2}}{{6}^{2/3}}=\frac{m\sqrt{5}}{n(2n{)}^{2/3}} \ \Rightarrow m=8,n=3 \ \Rightarrow 8m+12n=100 \end{gathered}$

Solutions · 1