Question

$\underset{x\to 2}{\lim }\frac{\sqrt{x+7}-3\sqrt{2x-3}}{\sqrt[3]{x+6}-2\sqrt[3]{3x-5}}=$

• A. $\frac{17}{23}$
• B. $\frac{34}{23}$
• C. $1$
• D. None of these

Answer 1

Answer: (B)

$\underset{x\to 2}{\lim }\frac{\sqrt{x+7}-3\sqrt{2x-3}}{\sqrt[3]{x+6}-2\sqrt[3]{3x-5}}$
$=\underset{x\to 2}{\lim }\frac{(x+7)-9(2x-3)}{\sqrt{x+7}+3\sqrt{2x-3}}\times [$ using $a-b=\frac{a^2-b^2}{a+b}]$
$\frac{(x+6{)}^{2/3}+2(x+6{)}^{1/3}(3x-5{)}^{1/3}+4(3x-5{)}^{2/3}}{(x+6)-8(3x-5)}$
$[$ using $a-b=\frac{a^3-b^3}{a^2+ab+b^2}]$
$=\underset{x\to z}{\lim }\frac{-17(x-2)}{\sqrt{x+7}+\sqrt[3]{2x-3}}\times \frac{(x+6{)}^{2/3}+2(x+6{)}^{1/3}(3x-5{)}^{1/3}+4(3x-5{)}^{2/3}}{-23(x-2)}$
$=\frac{-17}{\sqrt{9}+3\sqrt{1}}\cdot \frac{8^{2/3}+2.8^{1/3}+4}{-23}$
$=\frac{17}{6}\cdot \frac{12}{23}=\frac{34}{23}$