Statement -1 : `lim_(xrarralpha) sqrt(1-cos 2(x-alpha))/(x-alpha)` does not exist.
Statement-2 : `lim_(xrarr0) (|sin x|)/(x)` does not exist.

Options

A. Statement -1 is true, Statement-2 is true,, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for statement -1.
C. Statement-1 is true, Statement-2 is False.
D. Statement-1 is False, Statement-2 is true.

Explanation

To solve the problem, we need to analyze the two statements separately and determine their validity.

### Statement 1:
\[ \lim_{x \to \alpha} \frac{\sqrt{1 - \cos(2(x - \alpha))}}{x - \alpha} \text{ does not exist.} \]

**Step 1: Simplify the expression inside the limit.**

Using the identity \(1 - \cos(2\theta) = \sin^2(\theta)\), we can rewrite the expression as follows:
\[ 1 - \cos(2(x - \alpha)) = \sin^2(x - \alpha). \]

Thus, we can rewrite the limit:
\[ \lim_{x \to \alpha} \frac{\sqrt{\sin^2(x - \alpha)}}{x - \alpha}. \]

**Step 2: Simplify further.**

Since \(\sqrt{\sin^2(x - \alpha)} = |\sin(x - \alpha)|\), we have:
\[ \lim_{x \to \alpha} \frac{|\sin(x - \alpha)|}{x - \alpha}. \]

**Step 3: Analyze the limit.**

As \(x \to \alpha\), \(x - \alpha \to 0\). We can analyze the left-hand limit (LHL) and right-hand limit (RHL):

- For \(x \to \alpha^+\) (approaching from the right):
\[ \lim_{x \to \alpha^+} \frac{\sin(x - \alpha)}{x - \alpha} = 1. \]

- For \(x \to \alpha^-\) (approaching from the left):
\[ \lim_{x \to \alpha^-} \frac{-\sin(x - \alpha)}{x - \alpha} = -1. \]

Since the left-hand limit and right-hand limit are not equal, the limit does not exist.

### Conclusion for Statement 1:
The statement is **true**:
\[ \lim_{x \to \alpha} \frac{\sqrt{1 - \cos(2(x - \alpha))}}{x - \alpha} \text{ does not exist.} \]

---

### Statement 2:
\[ \lim_{x \to 0} \frac{|\sin x|}{x} \text{ does not exist.} \]

**Step 1: Analyze the limit.**

We can analyze this limit by considering the behavior of \(|\sin x|\) as \(x\) approaches \(0\):
- For \(x \to 0^+\):
\[ \lim_{x \to 0^+} \frac{\sin x}{x} = 1. \]

- For \(x \to 0^-\):
\[ \lim_{x \to 0^-} \frac{-\sin x}{x} = -1. \]

Since the left-hand limit and right-hand limit are not equal, the limit does not exist.

### Conclusion for Statement 2:
The statement is **true**:
\[ \lim_{x \to 0} \frac{|\sin x|}{x} \text{ does not exist.} \]

### Final Conclusion:
Both statements are true, and Statement 2 serves as a correct explanation for Statement 1.

---