Question

The value of ${lim}_{x\to \infty }\frac{2\sqrt{x}+3\sqrt[3]{3x}+5\sqrt[3]{3x}}{\sqrt{3x-2}+\sqrt[3]{32x-3}}$ is equal to $\frac{2}{\sqrt{a}}$ then find the value of a

• A. 3
• B. 5
• C. 7
• D. 9

Answer 1

The correct option is A 3

We have,

$\begin{array}{c}{lim}_{x\to \infty }\frac{2x^{\frac{1}{2}}+3x^{\frac{1}{3}}+5x^{\frac{1}{3}}}{{\left(3x-2\right)}^{\frac{1}{2}}+{\left(2x-3\right)}^{\frac{1}{3}}}\\ ={lim}_{x\to \infty }\frac{2+\frac{3x^{\frac{1}{3}}}{x^{\frac{1}{2}}}+5\frac{x^{\frac{1}{3}}}{x^{\frac{1}{2}}}}{{\left(\frac{3x-2}{x}\right)}^{\frac{1}{2}}+{\frac{\left(2x-3\right)}{x^{\frac{1}{2}}}}^{\frac{1}{3}}}\\ ={lim}_{x\to \infty }\frac{2+\frac{3}{x^{\frac{1}{6}}}+\frac{5}{x^{\frac{1}{6}}}}{{\left(3-\frac{2}{x}\right)}^{\frac{1}{2}}+{\frac{\left(2x-3\right)}{x^{\frac{1}{2}}}}^{\frac{1}{3}}}\\ =\frac{2}{\sqrt{3}}\\ \Rightarrow a=3\\ OptionAiscorrectanswer.\end{array}$