(\displaystyle\lim \_{x \rightarrow \infinity} \frac{(\sqrt{... | Mathematics | SolveeIt | Solveeit

Today, August 23

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

is equal to 27

Explanation

Solution

x→∞lim(x+x2−1)6+(x−x2−1)6(3x+1+3x−1)6+(3x+1−3x−1)6x3
x→∞limx3×⎩⎨⎧x6{(1+1−x21)6+(1−1−x21)6}x3{(3+x1+3−x1)6+(3+x1−3−x1)6}⎭⎬⎫
=26+0(23)6+0=33=(27)
So, the correct answer is (D) : is equal to 27