`lim_(x rarr oo)(((sqrt(3x+1)+sqrt(3x-1))^(6)+(sqrt(3x+1)-sqrt(3x-1))^6)/((x+sqrt(x^(2)-1))^(6)+(x-sqrt(x^(2)-1))^(6)))x^(3)`

Options

A. is equal to `(27)/(2)`
B. is equal to `9`
C. does not exist
D. is equal to `27`

Answer

To solve the limit

\[ \lim_{x \to \infty} \frac{(\sqrt{3x+1} + \sqrt{3x-1})^6 + (\sqrt{3x+1} - \sqrt{3x-1})^6}{(x + \sqrt{x^2 - 1})^6 + (x - \sqrt{x^2 - 1})^6} x^3, \]

we will simplify the expression step by step.

### Step 1: Simplify the Numerator

First, we analyze the terms in the numerator:

\[ \sqrt{3x+1} \text{ and } \sqrt{3x-1}. \]

As \(x\) approaches infinity, we can factor out \(\sqrt{x}\):

\[ \sqrt{3x+1} = \sqrt{x}\sqrt{3 + \frac{1}{x}} \quad \text{and} \quad \sqrt{3x-1} = \sqrt{x}\sqrt{3 - \frac{1}{x}}. \]

Thus, we have:

\[ \sqrt{3x+1} + \sqrt{3x-1} = \sqrt{x} \left( \sqrt{3 + \frac{1}{x}} + \sqrt{3 - \frac{1}{x}} \right). \]

As \(x \to \infty\), \(\sqrt{3 + \frac{1}{x}} \to \sqrt{3}\) and \(\sqrt{3 - \frac{1}{x}} \to \sqrt{3}\). Therefore:

\[ \sqrt{3x+1} + \sqrt{3x-1} \to 2\sqrt{3}\sqrt{x}. \]

Now, squaring this result gives:

\[ (\sqrt{3x+1} + \sqrt{3x-1})^6 \to (2\sqrt{3}\sqrt{x})^6 = 64 \cdot 3^3 \cdot x^3 = 64 \cdot 27 \cdot x^3 = 1728 x^3. \]

Next, we simplify the second term in the numerator:

\[ \sqrt{3x+1} - \sqrt{3x-1} = \sqrt{x} \left( \sqrt{3 + \frac{1}{x}} - \sqrt{3 - \frac{1}{x}} \right). \]

As \(x \to \infty\), \(\sqrt{3 + \frac{1}{x}} - \sqrt{3 - \frac{1}{x}} \to 0\). Thus, we find that:

\[ (\sqrt{3x+1} - \sqrt{3x-1})^6 \to 0. \]

So, the numerator simplifies to:

\[ 1728 x^3 + 0 = 1728 x^3. \]

### Step 2: Simplify the Denominator

Now we analyze the denominator:

\[ x + \sqrt{x^2 - 1} \text{ and } x - \sqrt{x^2 - 1}. \]

For large \(x\):

\[ \sqrt{x^2 - 1} = x\sqrt{1 - \frac{1}{x^2}} \to x. \]

Thus,

\[ x + \sqrt{x^2 - 1} \to x + x = 2x, \]
\[ x - \sqrt{x^2 - 1} \to x - x = 0. \]

Now, we can calculate:

\[ (x + \sqrt{x^2 - 1})^6 \to (2x)^6 = 64x^6, \]
\[ (x - \sqrt{x^2 - 1})^6 \to 0. \]

Thus, the denominator simplifies to:

\[ 64x^6 + 0 = 64x^6. \]

### Step 3: Combine and Simplify

Now we can combine the results:

\[ \lim_{x \to \infty} \frac{1728 x^3}{64 x^6} x^3 = \lim_{x \to \infty} \frac{1728}{64} \cdot \frac{x^3}{x^6} x^3 = \lim_{x \to \infty} \frac{1728}{64} \cdot \frac{x^6}{x^6} = \frac{1728}{64}. \]

Calculating \(\frac{1728}{64} = 27\).

### Final Answer

Thus, the limit is:

\[ \boxed{27}. \]