`lim_(xrarr oo) (logx^n-[x])/([x])` where `n in N and [.]` denotes the greatest integer function, is

Options

A. 1
B. -1
C. 0
D. none of these

Explanation

To solve the limit problem \( \lim_{x \to \infty} \frac{\log x^n - [x]}{[x]} \), where \( n \in \mathbb{N} \) and \([.]\) denotes the greatest integer function, we will follow these steps:

### Step 1: Rewrite the expression
We start by rewriting the logarithmic term:
\[ \log x^n = n \log x \]
Thus, the limit can be rewritten as:
\[ \lim_{x \to \infty} \frac{n \log x - [x]}{[x]} \]

### Step 2: Analyze the behavior of \([x]\)
As \( x \to \infty \), the greatest integer function \([x]\) approaches \( x \) because \([x] = x - \{x\}\), where \(\{x\}\) is the fractional part of \( x \). Therefore, we can approximate:
\[ [x] \approx x \]

### Step 3: Substitute the approximation into the limit
Now substituting \([x] \approx x\) into our limit gives:
\[ \lim_{x \to \infty} \frac{n \log x - x}{x} \]

### Step 4: Simplify the limit
This simplifies to:
\[ \lim_{x \to \infty} \left( \frac{n \log x}{x} - 1 \right) \]

### Step 5: Evaluate the limit of \(\frac{n \log x}{x}\)
To evaluate \(\lim_{x \to \infty} \frac{n \log x}{x}\), we can use L'Hôpital's rule since both the numerator and denominator approach infinity:
1. Differentiate the numerator: \( \frac{d}{dx}(n \log x) = \frac{n}{x} \)
2. Differentiate the denominator: \( \frac{d}{dx}(x) = 1 \)

Applying L'Hôpital's rule:
\[ \lim_{x \to \infty} \frac{n \log x}{x} = \lim_{x \to \infty} \frac{n/x}{1} = \lim_{x \to \infty} \frac{n}{x} = 0 \]

### Step 6: Final limit evaluation
Now substituting this back into our expression:
\[ \lim_{x \to \infty} \left( \frac{n \log x}{x} - 1 \right) = 0 - 1 = -1 \]

### Conclusion
Thus, the final result is:
\[ \lim_{x \to \infty} \frac{\log x^n - [x]}{[x]} = -1 \]