Question

Evaluate \[ \lim_{n\to\infty}\sqrt{2} \left[ \frac{(2+\sqrt{2})^n+(2-\sqrt{2})^n} {(2+\sqrt{2})^n-(2-\sqrt{2})^n} \right] \]

• A. \(2+\sqrt{2}\)
• B. \(2-\sqrt{2}\)
• C. \(1\)
• D. \(\sqrt{2}\)

Hint:

If \[ |r|\lt 1, \] then \[ r^n\to 0 \quad \text{as} \quad n\to\infty. \] In limits involving powers, divide by the dominant term to simplify the expression.

Answer 1

The Correct Option is D

Step 1: Observe the magnitudes of the terms.
We have \[ 2+\sqrt{2}\gt 1 \] and \[ 0\lt 2-\sqrt{2}\lt 1 \] Hence, \[ (2-\sqrt{2})^n\to 0 \quad \text{as} \quad n\to\infty \]

Step 2: Simplify the expression inside the bracket.
Consider \[ \frac{(2+\sqrt{2})^n+(2-\sqrt{2})^n} {(2+\sqrt{2})^n-(2-\sqrt{2})^n} \] Divide numerator and denominator by \[ (2+\sqrt{2})^n \] Then, \[ = \frac{ 1+\left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)^n }{ 1-\left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)^n } \] Now, \[ \left|\frac{2-\sqrt{2}}{2+\sqrt{2}}\right|\lt 1 \] Therefore, \[ \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)^n\to 0 \] Thus, \[ \frac{ 1+\left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)^n }{ 1-\left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)^n } \to 1 \]

Step 3: Compute the limit.
Hence, \[ \lim_{n\to\infty} \sqrt{2} \left[ \frac{(2+\sqrt{2})^n+(2-\sqrt{2})^n} {(2+\sqrt{2})^n-(2-\sqrt{2})^n} \right] = \sqrt{2}\times 1 \] \[ =\sqrt{2} \]

Step 4: Final conclusion.
Therefore, \[ \boxed{\sqrt{2}} \]