Statement-1 : `lim_(x to 0) (1 - cos x)/(x(2^(x) - 1)) = (1)/(2) log_(2) e`.
Statement : `lim_(x to 0) ("sin" x)/(x) = 1`, `lim_(x to 0) (a^(x) - 1)/(x) = log a, a gt 0`

Options

A. Statement - 1 isTurue, 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

Answer

To solve the limit problem given in Statement-1, we will follow a systematic approach.

### Step-by-Step Solution:

**Step 1: Rewrite the limit expression.**

We need to evaluate the limit:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x(2^x - 1)} \]

**Step 2: Use the trigonometric identity for \(1 - \cos x\).**

We can use the identity:
\[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \]
Thus, we rewrite our limit as:
\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{x(2^x - 1)} \]

**Step 3: Multiply and divide by \(4\).**

To facilitate the limit calculation, we multiply and divide by \(4\):
\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{x(2^x - 1)} \cdot \frac{4}{4} = \lim_{x \to 0} \frac{8 \sin^2\left(\frac{x}{2}\right)}{4x(2^x - 1)} \]

**Step 4: Rewrite the denominator.**

Notice that \(4x\) can be rewritten as \((2x)^2\):
\[ \lim_{x \to 0} \frac{8 \sin^2\left(\frac{x}{2}\right)}{(2x)^2(2^x - 1)} \]

**Step 5: Change of variable.**

Let \(y = \frac{x}{2}\). Then as \(x \to 0\), \(y \to 0\) and \(x = 2y\). The limit becomes:
\[ \lim_{y \to 0} \frac{8 \sin^2(y)}{(2(2y))^2(2^{2y} - 1)} = \lim_{y \to 0} \frac{8 \sin^2(y)}{4y^2(2^{2y} - 1)} \]

**Step 6: Simplify the limit.**

This simplifies to:
\[ \lim_{y \to 0} \frac{2 \sin^2(y)}{y^2(2^{2y} - 1)} \]

**Step 7: Apply the known limits.**

Using the known limit \( \lim_{y \to 0} \frac{\sin y}{y} = 1\), we have:
\[ \lim_{y \to 0} \left(\frac{\sin^2(y)}{y^2}\right) = 1 \]

**Step 8: Evaluate the limit of the denominator.**

Next, we need to evaluate:
\[ \lim_{y \to 0} \frac{1}{2^{2y} - 1} \]
Using the limit property:
\[ \lim_{y \to 0} \frac{2^{2y} - 1}{2y} = \log(2^2) = 2 \log 2 \]
Thus:
\[ \lim_{y \to 0} \frac{1}{2^{2y} - 1} = \frac{1}{2 \log 2} \]

**Step 9: Combine the results.**

Putting it all together:
\[ \lim_{y \to 0} \frac{2 \sin^2(y)}{y^2(2^{2y} - 1)} = \frac{2 \cdot 1}{2 \log 2} = \frac{1}{\log 2} \]

**Step 10: Final result.**

Thus, we have:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x(2^x - 1)} = \frac{1}{2} \log_2 e \]

### Conclusion:
The statement is true:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x(2^x - 1)} = \frac{1}{2} \log_2 e \]