The value of `lim_(xrarr0^(+))((1-cos(sin^(2)x))/(x^(2)))^((log_(e)(1-2x^(2)))/(sin^(2)x))` is

Options

A. 0
B. e
C. `-1`
D. `oo`

Explanation

To solve the limit problem, we will follow a systematic approach to evaluate the limit step by step.

**Given:**
\[ L = \lim_{x \to 0^+} \left( \frac{1 - \cos(\sin^2 x)}{x^2} \right)^{\frac{\log_e(1 - 2x^2)}{\sin^2 x}} \]

### Step 1: Substitute \( x = h \) where \( h \to 0^+ \)
We rewrite the limit as:
\[ L = \lim_{h \to 0^+} \left( \frac{1 - \cos(\sin^2 h)}{h^2} \right)^{\frac{\log_e(1 - 2h^2)}{\sin^2 h}} \]

### Step 2: Analyze the numerator \( 1 - \cos(\sin^2 h) \)
Using the trigonometric identity \( 1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right) \):
\[ 1 - \cos(\sin^2 h) = 2 \sin^2\left(\frac{\sin^2 h}{2}\right) \]

### Step 3: Simplify \( \sin^2\left(\frac{\sin^2 h}{2}\right) \)
As \( h \to 0 \), \( \sin h \approx h \):
\[ \sin^2 h \approx h^2 \implies \frac{\sin^2 h}{2} \approx \frac{h^2}{2} \]
Thus,
\[ \sin^2\left(\frac{\sin^2 h}{2}\right) \approx \sin^2\left(\frac{h^2}{2}\right) \approx \left(\frac{h^2}{2}\right)^2 = \frac{h^4}{4} \]

### Step 4: Substitute back into the limit
Now we have:
\[ 1 - \cos(\sin^2 h) \approx 2 \cdot \frac{h^4}{4} = \frac{h^4}{2} \]
So,
\[ \frac{1 - \cos(\sin^2 h)}{h^2} \approx \frac{\frac{h^4}{2}}{h^2} = \frac{h^2}{2} \]

### Step 5: Evaluate the logarithmic part
Next, we need to evaluate:
\[ \log_e(1 - 2h^2) \approx -2h^2 \quad \text{(as \( h \to 0 \))} \]
Thus,
\[ \frac{\log_e(1 - 2h^2)}{\sin^2 h} \approx \frac{-2h^2}{h^2} = -2 \]

### Step 6: Combine results
Now substituting back into the limit:
\[ L = \lim_{h \to 0^+} \left( \frac{h^2}{2} \right)^{-2} \]
This simplifies to:
\[ L = \lim_{h \to 0^+} \left( \frac{2}{h^2} \right)^{2} = \lim_{h \to 0^+} \frac{4}{h^4} \]

### Step 7: Evaluate the limit
As \( h \to 0^+ \):
\[ \frac{4}{h^4} \to \infty \]

### Conclusion
Thus, the final answer is:
\[ L = \infty \]