### Limits Question 5

#### Question 5 - 31 January - Shift 2

Hence $y \in(-\infty, \frac{-21}{4}] \cup[0, \infty)$

$\lim _{x \to \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^{6}+(\sqrt{3 x+1}-\sqrt{3 x-1})^{6}}{(x+\sqrt{x^{2}-1})^{6}+(x-\sqrt{x^{2}-1})^{6}} x^{3}$

(1) is equal to 9

(2) is equal to 27

(3) does not exist

(4) is equal to $\frac{27}{2}$

## Show Answer

#### Answer: (2)

#### Solution:

#### Formula: Standard result: limit of 1/x as x goes to infinity is 0

$ \begin{aligned} & \lim _{x \to \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^{6}+(\sqrt{3 x+1}-\sqrt{3 x-1})^{6}}{(x+\sqrt{x^{2}-1})^{6}+(x-\sqrt{x^{2}-1}){6}} x^{3} \\ & \lim _{x \to \infty} x^{3} \times{\frac{x^{3}{(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}})^{6}+(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}})^{6}}}{x^{6}{(1+\sqrt{1-\frac{1}{x^{2}}})^{6}+(1-\sqrt{1-\frac{1}{x^{2}}})^{6}}}} \\ & =\frac{(2 \sqrt{3})^{6}+0}{2^{6}+0}=3^{3} = 27 \end{aligned} $