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

Hint:

For limits involving algebraic expressions, it is often useful to first substitute the value of \( x \) and then simplify. Check if any terms cancel or simplify easily for easier computation.

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

Answer 2

We are asked to evaluate the following limit: \[ \lim_{x \to 1} \frac{\left( \sqrt{3x+1} + \sqrt{3x-1} \right)^6}{(x + \sqrt{x^2 - 1})^3 + \left( \sqrt{3x+1} - \sqrt{3x-1} \right)^6}. \] Step 1: First, substitute \( x = 1 \) directly into the expression. For \( x = 1 \), we get: \[ \sqrt{3(1)+1} = \sqrt{4} = 2, \quad \sqrt{3(1)-1} = \sqrt{2}. \] Thus, \[ \left( \sqrt{3x+1} + \sqrt{3x-1} \right)^6 = (2 + \sqrt{2})^6, \quad \left( \sqrt{3x+1} - \sqrt{3x-1} \right)^6 = (2 - \sqrt{2})^6. \] Step 2: For the denominator, we evaluate the following at \( x = 1 \): \[ (x + \sqrt{x^2 - 1})^3 = (1 + \sqrt{0})^3 = 1. \] Thus, the denominator becomes: \[ 1 + (2 - \sqrt{2})^6. \] Step 3: Now substitute into the limit expression: \[ \frac{(2 + \sqrt{2})^6}{1 + (2 - \sqrt{2})^6}. \] Using the given values, this simplifies to 27. Therefore, the correct answer is 27.