If `a_(n)` and `b_(n)` are positive integers and `a_(n)+sqrt2b_(n)=(2+sqrt2))^(n)`, then `lim_(nrarroo) ((a_(n))/(b_(n)))`=

Options

A. `sqrt2`
B. 2
C. `e^(sqrt2)`
D. `e^(2)`

Explanation

To solve the problem, we start with the given equation:

\[ a_n + \sqrt{2} b_n = (2 + \sqrt{2})^n \]

We need to find the limit:

\[ \lim_{n \to \infty} \frac{a_n}{b_n} \]

### Step 1: Express \( a_n \) and \( b_n \)

We can express \( a_n \) and \( b_n \) in terms of \( (2 + \sqrt{2})^n \) and \( (2 - \sqrt{2})^n \).

Using the binomial theorem, we can expand \( (2 + \sqrt{2})^n \) and \( (2 - \sqrt{2})^n \):

\[ (2 + \sqrt{2})^n + (2 - \sqrt{2})^n = 2 \cdot 2^n \quad \text{(since \( (2 - \sqrt{2})^n \) approaches 0 as \( n \to \infty \))} \]

Thus, we can write:

\[ a_n = \frac{(2 + \sqrt{2})^n + (2 - \sqrt{2})^n}{2} \]
\[ b_n = \frac{(2 + \sqrt{2})^n - (2 - \sqrt{2})^n}{2\sqrt{2}} \]

### Step 2: Simplify \( \frac{a_n}{b_n} \)

Now, we can find \( \frac{a_n}{b_n} \):

\[ \frac{a_n}{b_n} = \frac{\frac{(2 + \sqrt{2})^n + (2 - \sqrt{2})^n}{2}}{\frac{(2 + \sqrt{2})^n - (2 - \sqrt{2})^n}{2\sqrt{2}}} \]

This simplifies to:

\[ \frac{a_n}{b_n} = \sqrt{2} \cdot \frac{(2 + \sqrt{2})^n + (2 - \sqrt{2})^n}{(2 + \sqrt{2})^n - (2 - \sqrt{2})^n} \]

### Step 3: Analyze the limit as \( n \to \infty \)

As \( n \to \infty \), the term \( (2 - \sqrt{2})^n \) approaches 0 because \( 2 - \sqrt{2} < 1 \). Therefore, we can approximate:

\[ \frac{a_n}{b_n} \approx \sqrt{2} \cdot \frac{(2 + \sqrt{2})^n}{(2 + \sqrt{2})^n} = \sqrt{2} \]

### Step 4: Find the limit

Thus, we find that:

\[ \lim_{n \to \infty} \frac{a_n}{b_n} = \sqrt{2} \]

### Conclusion

The final answer is:

\[ \lim_{n \to \infty} \frac{a_n}{b_n} = \sqrt{2} \]