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

We need to evaluate the limit:

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

First, we simplify the expressions inside the limit. As \(x\) approaches infinity, we approximate the square roots:

• \(\sqrt{3x+1}\approx \sqrt{3x}=\sqrt{3}\sqrt{x}\)
• \(\sqrt{3x-1}\approx \sqrt{3x}=\sqrt{3}\sqrt{x}\)

Thus:

• \(\sqrt{3x+1}+\sqrt{3x-1} \approx 2\sqrt{3}\sqrt{x}\)
• \(\sqrt{3x+1}-\sqrt{3x-1} \approx 0\)

Then the numerator becomes:

$$(2\sqrt{3}\sqrt{x})^6$$

Since \((a-b)^6\) will result in 0 when \(a=b\), the term \((\sqrt{3x+1}-\sqrt{3x-1})^6 \approx 0\). Thus, numerator simplifies to:
\((2\sqrt{3}\sqrt{x})^6=(2^6)(3^3)(x^3)=64\cdot27\cdot x^3=1728x^3\)

Expression	Approximation
\(x+\sqrt{x^2-1}\)	\(2x\)
\(x-\sqrt{x^2-1}\)	\(0\)

Now consider the denominator:

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

Plugging into the limit expression, we get:
$$\lim_{x\to\infty}\frac{1728x^3}{64x^6}x^3=\lim_{x\to\infty}\frac{1728}{64}=\lim_{x\to\infty}27$$
Thus, the eigenvalue simplifies to 27. The answer is: 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.