` underset( x to 0 ) lim (sin ^(2) 2x)/( sin ^(2) 4x) ` is equal to

Options

A. `(1)/(2)`
B. ` (1)/(4)`
C. ` (1)/(8)`
D. ` (1)/(16)`

Answer

To solve the limit \( \lim_{x \to 0} \frac{\sin^2(2x)}{\sin^2(4x)} \), we can proceed step by step as follows:

### Step 1: Rewrite the limit
We start with the expression:
\[ \lim_{x \to 0} \frac{\sin^2(2x)}{\sin^2(4x)} \]

### Step 2: Use the identity for sine
We can use the identity \( \sin(kx) \approx kx \) as \( x \to 0 \). Thus, we rewrite the sine functions:
\[ \sin(2x) \approx 2x \quad \text{and} \quad \sin(4x) \approx 4x \]

### Step 3: Substitute the approximations
Substituting these approximations into our limit gives:
\[ \lim_{x \to 0} \frac{(2x)^2}{(4x)^2} \]

### Step 4: Simplify the expression
This simplifies to:
\[ \lim_{x \to 0} \frac{4x^2}{16x^2} \]
We can cancel \( x^2 \) from the numerator and denominator (as long as \( x \neq 0 \)):
\[ \lim_{x \to 0} \frac{4}{16} \]

### Step 5: Final simplification
This further simplifies to:
\[ \frac{4}{16} = \frac{1}{4} \]

### Conclusion
Thus, we find that:
\[ \lim_{x \to 0} \frac{\sin^2(2x)}{\sin^2(4x)} = \frac{1}{4} \]

### Final Answer
The limit is \( \frac{1}{4} \).

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