STATEMENT-1 : `lim_(x->oo)(log[x])/([x])=0`. STATEMENT-2 : `lim_(x->0)(sqrt(sec^2-1))/x` does not exist. STATEMENT-3: `lim_(x->2)(x-1)^(1/(x-2)) = 1`

Options

A. TTT
B. TTF
C. FTF
D. FFF

Answer

To solve the given statements, we will analyze each statement one by one.

### Statement 1:
**Evaluate**: \(\lim_{x \to \infty} \frac{\log[x]}{[x]}\)

1. **Understanding the components**:
- \(\log[x]\) refers to the logarithm of the greatest integer function of \(x\), which is denoted as \(\log(\lfloor x \rfloor)\).
- \([x]\) is the greatest integer less than or equal to \(x\), denoted as \(\lfloor x \rfloor\).

2. **Behavior as \(x \to \infty\)**:
- As \(x\) approaches infinity, \(\lfloor x \rfloor\) also approaches infinity.
- Thus, both the numerator and denominator approach infinity.

3. **Applying L'Hôpital's Rule**:
- Since both the numerator and denominator approach infinity, we can apply L'Hôpital's Rule:
\[ \lim_{x \to \infty} \frac{\log(\lfloor x \rfloor)}{\lfloor x \rfloor} = \lim_{x \to \infty} \frac{\frac{d}{dx}(\log(\lfloor x \rfloor))}{\frac{d}{dx}(\lfloor x \rfloor)} \]
- The derivative of \(\lfloor x \rfloor\) is 0 almost everywhere, but we can consider the behavior of \(\log(\lfloor x \rfloor)\) as \(x\) becomes very large.

4. **Conclusion**:
- Since \(\log(\lfloor x \rfloor)\) grows much slower than \(\lfloor x \rfloor\), we find that:
\[ \lim_{x \to \infty} \frac{\log(\lfloor x \rfloor)}{\lfloor x \rfloor} = 0 \]
- Therefore, **Statement 1 is true**.

### Statement 2:
**Evaluate**: \(\lim_{x \to 0} \frac{\sqrt{\sec^2 x - 1}}{x}\)

1. **Simplifying the expression**:
- We know that \(\sec^2 x - 1 = \tan^2 x\), hence:
\[ \sqrt{\sec^2 x - 1} = \sqrt{\tan^2 x} = |\tan x| \]

2. **Limit as \(x\) approaches 0**:
- Thus, the limit becomes:
\[ \lim_{x \to 0} \frac{|\tan x|}{x} \]
- As \(x\) approaches 0, \(\tan x \approx x\), so:
\[ \lim_{x \to 0} \frac{|\tan x|}{x} = \lim_{x \to 0} \frac{x}{x} = 1 \]

3. **Conclusion**:
- Since both the left-hand limit and right-hand limit yield the same result, the limit exists.
- Therefore, **Statement 2 is false**.

### Statement 3:
**Evaluate**: \(\lim_{x \to 2} (x-1)^{\frac{1}{x-2}}\)

1. **Rewriting the expression**:
- We can rewrite the limit as:
\[ \lim_{x \to 2} e^{\frac{\log(x-1)}{x-2}} \]

2. **Finding the limit of the exponent**:
- As \(x\) approaches 2, both \(\log(x-1)\) approaches \(\log(1) = 0\) and \(x-2\) approaches 0.
- This is an indeterminate form \(\frac{0}{0}\).

3. **Applying L'Hôpital's Rule**:
- We apply L'Hôpital's Rule:
\[ \lim_{x \to 2} \frac{\log(x-1)}{x-2} = \lim_{x \to 2} \frac{\frac{1}{x-1}}{1} = \frac{1}{2-1} = 1 \]

4. **Conclusion**:
- Therefore, the limit becomes:
\[ e^1 = e \]
- Hence, **Statement 3 is false**.

### Final Summary:
- **Statement 1**: True
- **Statement 2**: False
- **Statement 3**: False