Question

$\displaystyle\lim _{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3$

• A. is equal to $\frac{27}{2}$
• B. is equal to 9
• C. does not exist
• D. is equal to 27

Answer 1

The Correct Option is D

$\underset{x\to \infty }{\lim }\frac{(\sqrt{3x+1}+\sqrt{3x-1}{)}^6+(\sqrt{3x+1}-\sqrt{3x-1}{)}^6}{{\left(x+\sqrt{x^2-1}\right)}^6+{\left(x-\sqrt{x^2-1}\right)}^6}{x}^3$
$\underset{x\to \infty }{\lim }{x}^3\times \left\{\frac{x^3\left\{{\left(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}\right)}^6+{\left(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}\right)}^6\right\}}{x^6\left\{{\left(1+\sqrt{1-\frac{1}{x^2}}\right)}^6+{\left(1-\sqrt{1-\frac{1}{x^2}}\right)}^6\right\}}\right\}$
$=\frac{(2\sqrt{3}{)}^6+0}{2^6+0}=3^3=(27)$
So, the correct answer is (D) : is equal to 27