Question

$\displaystyle\lim _{x \rightarrow 2} \frac{\sqrt{x+7}-3 \sqrt{2 x-3}}{\sqrt[3]{x+6}-2 \sqrt[3]{3 x-5}}=$

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

Answer 1

Answer: (B)

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