Statement -1 : `lim_( x to 0) (sin x)/x ` exists ,.
Statement -2 : |x| is differentiable at x=0
Statement -3 : If `lim_(x to 0) (tan kx)/(sin 5x) = 3` , then k = 15

Options

A. T F T
B. TTT
C. FFF
D. FFT

Answer

To determine the validity of the given statements, we will analyze each statement step by step.

### Step 1: Analyze Statement 1
**Statement 1:** \( \lim_{x \to 0} \frac{\sin x}{x} \) exists.

**Solution:**
The limit \( \lim_{x \to 0} \frac{\sin x}{x} \) is a well-known limit in calculus. We can evaluate it using the Squeeze Theorem or L'Hôpital's Rule.

Using the Squeeze Theorem:
- We know that \( \sin x \) is bounded by \( x \) and \( -x \) for small values of \( x \).
- Therefore, \( -1 \leq \frac{\sin x}{x} \leq 1 \).
- As \( x \) approaches 0, both bounds approach 1.
- Hence, by the Squeeze Theorem, \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).

Thus, **Statement 1 is true.**

### Step 2: Analyze Statement 2
**Statement 2:** \( |x| \) is differentiable at \( x = 0 \).

**Solution:**
To check the differentiability of \( |x| \) at \( x = 0 \), we need to find the derivative from both sides.

- For \( x > 0 \), \( |x| = x \) and the derivative \( f'(x) = 1 \).
- For \( x < 0 \), \( |x| = -x \) and the derivative \( f'(x) = -1 \).
- At \( x = 0 \), the left-hand derivative is -1 and the right-hand derivative is 1.

Since the left-hand and right-hand derivatives are not equal, \( |x| \) is not differentiable at \( x = 0 \).

Thus, **Statement 2 is false.**

### Step 3: Analyze Statement 3
**Statement 3:** If \( \lim_{x \to 0} \frac{\tan(kx)}{\sin(5x)} = 3 \), then \( k = 15 \).

**Solution:**
We start with the limit:
\[ \lim_{x \to 0} \frac{\tan(kx)}{\sin(5x)}. \]
Both \( \tan(kx) \) and \( \sin(5x) \) approach 0 as \( x \to 0 \), so we can apply L'Hôpital's Rule or use the small-angle approximations.

Using the small-angle approximations:
- \( \tan(kx) \approx kx \) as \( x \to 0 \).
- \( \sin(5x) \approx 5x \) as \( x \to 0 \).

Thus, we can rewrite the limit:
\[ \lim_{x \to 0} \frac{\tan(kx)}{\sin(5x)} = \lim_{x \to 0} \frac{kx}{5x} = \frac{k}{5}. \]
Setting this equal to 3 gives:
\[ \frac{k}{5} = 3 \implies k = 15. \]

Thus, **Statement 3 is true.**

### Conclusion
- **Statement 1:** True
- **Statement 2:** False
- **Statement 3:** True