If `lim_(xrarr0) (f(x))/(x^(2))=a and lim_(xrarr0) (f(1-cosx))/(g(x)sin^(2)x)=b` (where `b ne 0`),
then `lim_(xrarr0) (g(1-cos2x))/(x^(4))` is

Options

A. `(4a)/(b)`
B. `(a)/(4b)`
C. `(a)/(b)`
D. none of these

Explanation

To solve the problem step by step, we will analyze the given limits and derive the required limit.

### Given:
1. \(\lim_{x \to 0} \frac{f(x)}{x^2} = a\)
2. \(\lim_{x \to 0} \frac{f(1 - \cos x)}{g(x) \sin^2 x} = b\) (where \(b \neq 0\))

We need to find:
\[ \lim_{x \to 0} \frac{g(1 - \cos 2x)}{x^4} \]

### Step 1: Rewrite \(1 - \cos x\)
Using the trigonometric identity:
\[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \]
We can express \(1 - \cos 2x\) as:
\[ 1 - \cos 2x = 2 \sin^2(x) \]

### Step 2: Substitute into the limit
Now we can rewrite the limit we want to find:
\[ \lim_{x \to 0} \frac{g(1 - \cos 2x)}{x^4} = \lim_{x \to 0} \frac{g(2 \sin^2 x)}{x^4} \]

### Step 3: Use the second limit
From the second limit, we have:
\[ \lim_{x \to 0} \frac{f(1 - \cos x)}{g(x) \sin^2 x} = b \]
Substituting \(1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)\):
\[ \lim_{x \to 0} \frac{f(2 \sin^2\left(\frac{x}{2}\right))}{g(x) \sin^2 x} = b \]

### Step 4: Analyze the limit
As \(x \to 0\), we have \(f(2 \sin^2\left(\frac{x}{2}\right))\) behaving like \(f(0)\) since \(\sin^2\left(\frac{x}{2}\right) \to 0\). Thus, we can write:
\[ f(2 \sin^2\left(\frac{x}{2}\right)) \sim 2a \sin^2\left(\frac{x}{2}\right) \quad \text{(from the first limit)} \]

### Step 5: Substitute back into the limit
Now substituting back:
\[ \lim_{x \to 0} \frac{2a \sin^2\left(\frac{x}{2}\right)}{g(x) \sin^2 x} = b \]

### Step 6: Simplify \(\sin^2 x\)
Using the small angle approximation:
\[ \sin x \sim x \quad \text{and} \quad \sin^2 x \sim x^2 \]
Thus:
\[ \sin^2\left(\frac{x}{2}\right) \sim \left(\frac{x}{2}\right)^2 = \frac{x^2}{4} \]
So:
\[ \sin^2 x \sim x^2 \]

### Step 7: Final limit
Now we can substitute:
\[ \frac{2a \cdot \frac{x^2}{4}}{g(x) \cdot x^2} = b \implies \frac{a}{2g(x)} = b \]
Thus:
\[ g(x) \sim \frac{a}{2b} \]

### Step 8: Find the required limit
Now we can substitute back into our original limit:
\[ \lim_{x \to 0} \frac{g(2 \sin^2 x)}{x^4} = \lim_{x \to 0} \frac{g(2 \cdot \frac{x^2}{4})}{x^4} = \lim_{x \to 0} \frac{g\left(\frac{x^2}{2}\right)}{x^4} \]
As \(x \to 0\), \(g\left(\frac{x^2}{2}\right) \sim \frac{a}{2b}\):
\[ \lim_{x \to 0} \frac{\frac{a}{2b}}{x^4} = \frac{a}{2b} \cdot \frac{1}{x^4} \]

### Conclusion
Thus, the final limit is:
\[ \lim_{x \to 0} \frac{g(1 - \cos 2x)}{x^4} = \frac{4a}{b} \]

### Final Answer:
The limit is \(\frac{4a}{b}\).