Let [x] denote the greatest integer less than or equal to x for any real number x. then `lim_(xrarroo)([nsqrt2])/(n)` is equal to

Options

A. 0
B. 2
C. `sqrt2`
D. 1

Answer

To solve the limit \( \lim_{n \to \infty} \frac{[\sqrt{2}^n]}{n} \), where \([\cdot]\) denotes the greatest integer function, we can follow these steps:

### Step 1: Understand the Expression
The expression we need to evaluate is \( \frac{[\sqrt{2}^n]}{n} \). Here, \([\sqrt{2}^n]\) is the greatest integer less than or equal to \(\sqrt{2}^n\).

### Step 2: Analyze \(\sqrt{2}^n\)
As \(n\) approaches infinity, \(\sqrt{2}^n\) grows exponentially. Specifically, we know that:
\[ \sqrt{2}^n = 2^{n/2} \]
Thus, as \(n\) increases, \(\sqrt{2}^n\) becomes very large.

### Step 3: Apply the Greatest Integer Function
The greatest integer function \([\sqrt{2}^n]\) can be approximated as:
\[ [\sqrt{2}^n] \approx \sqrt{2}^n \quad \text{(as \(n\) becomes very large)} \]
This means that:
\[ [\sqrt{2}^n] = \sqrt{2}^n - \epsilon_n \]
where \(0 \leq \epsilon_n < 1\).

### Step 4: Substitute into the Limit
Now we can substitute this approximation into our limit:
\[ \lim_{n \to \infty} \frac{[\sqrt{2}^n]}{n} = \lim_{n \to \infty} \frac{\sqrt{2}^n - \epsilon_n}{n} \]
This can be simplified to:
\[ \lim_{n \to \infty} \left( \frac{\sqrt{2}^n}{n} - \frac{\epsilon_n}{n} \right) \]

### Step 5: Evaluate Each Part of the Limit
1. **Evaluate \(\lim_{n \to \infty} \frac{\sqrt{2}^n}{n}\)**:
- Since \(\sqrt{2}^n\) grows exponentially while \(n\) grows linearly, this limit approaches infinity:
\[ \lim_{n \to \infty} \frac{\sqrt{2}^n}{n} = \infty \]

2. **Evaluate \(\lim_{n \to \infty} \frac{\epsilon_n}{n}\)**:
- Since \(\epsilon_n\) is bounded (it is less than 1), this limit approaches 0:
\[ \lim_{n \to \infty} \frac{\epsilon_n}{n} = 0 \]

### Step 6: Combine the Results
Putting it all together, we have:
\[ \lim_{n \to \infty} \left( \frac{\sqrt{2}^n}{n} - \frac{\epsilon_n}{n} \right) = \infty - 0 = \infty \]

### Conclusion
Thus, the limit is:
\[ \lim_{n \to \infty} \frac{[\sqrt{2}^n]}{n} = \infty \]