Question

Evaluate $L=\underset{x\to 2}{\lim }$$\frac{\sqrt{\left(x+7\right)}-3\sqrt{\left(2x+3\right)}}{\sqrt[3]{\left(x+6\right)}-2\sqrt[3]{\left(3x-5\right)}}$

• A. $\frac{17}{9}$
• B. $\frac{17}{18}$
• C. $\frac{34}{23}$
• D. $\frac{26}{7}$

Answer 1

Answer: (C)

We have
$L=\underset{x\to 2}{\lim }$$\frac{\sqrt{\left(x+7\right)}-3\sqrt{\left(2x-3\right)}}{\sqrt[3]{\left(x+6\right)}-2\sqrt[3]{\left(3x-5\right)}}\left(\frac{0}{0}form\right)$
Let $x-2=t$ such that when $x\to 2,t\to 0.$ Then
$L=\underset{t\to 0}{\lim }$ $\frac{{\left(t+9\right)}^{\frac{1}{2}}-3{\left(2t+1\right)}^{\frac{1}{2}}}{{\left(t+8\right)}^{\frac{1}{3}}-2{\left(3t+1\right)}^{\frac{1}{3}}}\left(\frac{0}{0}form\right)$
$=\frac{3}{2}$ $\underset{t\to 0}{\lim }$ $\frac{{\left(1+\frac{t}{9}\right)}^{\frac{1}{2}}-{\left(2t+1\right)}^{\frac{1}{2}}}{{\left(1+\frac{t}{8}\right)}^{\frac{1}{3}}-{\left(3t+1\right)}^{\frac{1}{3}}}\left(\frac{0}{0}form\right)$
$=\frac{3}{2}\underset{t\to 0}{\lim }$$\frac{\frac{1}{2}\frac{t}{9}-\left(2t\right)\frac{1}{2}}{\frac{t}{8}\frac{1}{3}-\left(3t\right)\frac{1}{3}}=\frac{3}{2}\frac{\frac{1}{18}-1}{\frac{1}{24}-1}=\frac{34}{23}.$