Question

If \[ \lim\limits_{x \to \infty} \frac{\left(\sqrt{2x+1} + \sqrt{2x-1}\right) + \left(\sqrt{2x+1} - \sqrt{2x-1}\right) P x^4 - 16} {(x+\sqrt{x^2 - 2}) + (x - \sqrt{x^2 - 2})} = 1, \] then P = ?

• A. \( 16 \)
• B. \( 64 \)
• C. \( \frac{1}{64} \)
• D. \( \frac{1}{16} \)

Hint:

For large \( x \), use first-order approximations \( \sqrt{a \pm b} \approx \sqrt{a} \pm \frac{b}{2\sqrt{a}} \) to simplify square root expressions.

Answer 1

The Correct Option is D

Step 1: Simplifying the denominator
The denominator consists of: \[ (x+\sqrt{x^2 - 2}) + (x - \sqrt{x^2 - 2}). \] Since \( x + \sqrt{x^2 - 2} \) and \( x - \sqrt{x^2 - 2} \) are conjugates, their sum simplifies to: \[ (x+\sqrt{x^2 - 2}) + (x - \sqrt{x^2 - 2}) = 2x. \] Step 2: Simplifying the numerator
Rewriting the terms: \[ \left(\sqrt{2x+1} + \sqrt{2x-1}\right) + \left(\sqrt{2x+1} - \sqrt{2x-1}\right) P x^4 - 16. \] Using the identity: \[ \sqrt{a} + \sqrt{b} = \frac{a-b}{\sqrt{a} - \sqrt{b}}. \] Approximating for large \( x \): \[ \sqrt{2x+1} \approx \sqrt{2x} + \frac{1}{2\sqrt{2x}}, \quad \sqrt{2x-1} \approx \sqrt{2x} - \frac{1}{2\sqrt{2x}}. \] Thus, \[ \sqrt{2x+1} + \sqrt{2x-1} = 2\sqrt{2x}. \] Similarly, \[ \sqrt{2x+1} - \sqrt{2x-1} = \frac{1}{\sqrt{2x}}. \] Step 3: Evaluating the limit
\[ \lim\limits_{x \to \infty} \frac{(2\sqrt{2x}) + \left(\frac{1}{\sqrt{2x}}\right) P x^4 - 16}{2x} = 1. \] For large \( x \), dominant terms in the numerator and denominator must balance. Comparing coefficients, \[ P = \frac{1}{16}. \]