`lim_(xrarr oo) (log[x])/(x)` , where `[x]` denotes the greatest integer less than or equal to x, is

Options

A. 0
B. 1
C. -1
D. non-existent

Explanation

To solve the limit \( \lim_{x \to \infty} \frac{\log[x]}{x} \), where \([x]\) denotes the greatest integer less than or equal to \(x\), we can follow these steps:

### Step 1: Understanding the Greatest Integer Function
The greatest integer function \([x]\) gives us the largest integer less than or equal to \(x\). For large values of \(x\), we can express \(x\) as \(n + k\), where \(n\) is an integer and \(k\) is a small fractional part such that \(0 \leq k < 1\). Thus, we have:
\[ [x] = n \]

### Step 2: Rewrite the Limit
Now we can rewrite the limit:
\[ \lim_{x \to \infty} \frac{\log[x]}{x} = \lim_{n \to \infty} \frac{\log n}{n + k} \]
As \(x\) approaches infinity, \(n\) also approaches infinity, and \(k\) becomes negligible.

### Step 3: Analyze the Form of the Limit
The expression \(\frac{\log n}{n + k}\) approaches the form \(\frac{\infty}{\infty}\) as \(n\) approaches infinity. This allows us to use L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule
Using L'Hôpital's Rule, we differentiate the numerator and the denominator:
- The derivative of \(\log n\) is \(\frac{1}{n}\).
- The derivative of \(n + k\) is \(1\) (since \(k\) is a constant with respect to \(n\)).

Thus, we have:
\[ \lim_{n \to \infty} \frac{\log n}{n + k} = \lim_{n \to \infty} \frac{\frac{1}{n}}{1} = \lim_{n \to \infty} \frac{1}{n} \]

### Step 5: Evaluate the Limit
As \(n\) approaches infinity, \(\frac{1}{n}\) approaches \(0\):
\[ \lim_{n \to \infty} \frac{1}{n} = 0 \]

### Conclusion
Therefore, the limit is:
\[ \lim_{x \to \infty} \frac{\log[x]}{x} = 0 \]