Question

Evaluate $L=\displaystyle \lim_{x \to 2}$$\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=\displaystyle \lim_{x \to 2}$$\frac{\sqrt{\left(x+7\right)}-3\sqrt{\left(2x-3\right)}}{\sqrt[3]{\left(x+6\right)}-2\sqrt[3]{\left(3x-5\right)}} \left(\frac{0}{0}form\right)$
Let $x - 2 = t$ such that when $x \rightarrow 2, t \rightarrow 0.$ Then
$L=\displaystyle \lim_{t \to 0}$ $\frac{\left(t+9\right)^{\frac{1}{2}}-3\left(2t+1\right)^{\frac{1}{2}}}{\left(t+8\right)^{\frac{1}{3}}-2\left(3t+1\right)^{\frac{1}{3}}}\,\left(\frac{0}{0} form\right)$
$=\frac{3}{2}$ $\displaystyle \lim_{t \to 0}$ $\frac{\left(1+\frac{t}{9}\right)^{\frac{1}{2}}-\left(2t+1\right)^{\frac{1}{2}}}{\left(1+\frac{t}{8}\right)^{\frac{1}{3}}-\left(3t+1\right)^{\frac{1}{3}}}\, \left(\frac{0}{0} form\right)$
$=\frac{3}{2} \displaystyle \lim_{t \to 0}$$\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}.$