`lim _(xto0) (""^(3)sqrt(1+sin ^(2)x )- ""^(4) sqrt(1-2 tan x))/(sin x+ tan ^(2) x)` is equal to:

Options

A. `-1`
B. 1
C. `1/2`
D. `-1/2`

Answer

To solve the limit

\[ \lim_{x \to 0} \frac{\sqrt[3]{1 + \sin^2 x} - \sqrt[4]{1 - 2 \tan x}}{\sin x + \tan^2 x}, \]

we will follow these steps:

### Step 1: Identify the form of the limit
As \( x \to 0 \), we can evaluate the numerator and denominator separately:
- The numerator:
- \( \sin^2 x \to 0 \) implies \( \sqrt[3]{1 + \sin^2 x} \to \sqrt[3]{1} = 1 \)
- \( \tan x \to 0 \) implies \( \sqrt[4]{1 - 2 \tan x} \to \sqrt[4]{1} = 1 \)
- Therefore, the numerator \( \sqrt[3]{1 + \sin^2 x} - \sqrt[4]{1 - 2 \tan x} \to 1 - 1 = 0 \).

- The denominator:
- \( \sin x \to 0 \) and \( \tan^2 x \to 0 \) implies \( \sin x + \tan^2 x \to 0 + 0 = 0 \).

Thus, we have a \( \frac{0}{0} \) indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have the \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator.

### Step 3: Differentiate the numerator
Let \( u = \sqrt[3]{1 + \sin^2 x} \) and \( v = \sqrt[4]{1 - 2 \tan x} \).

Using the chain rule:
- The derivative of \( u \):
\[ \frac{du}{dx} = \frac{1}{3}(1 + \sin^2 x)^{-\frac{2}{3}} \cdot 2 \sin x \cos x = \frac{2 \sin x \cos x}{3(1 + \sin^2 x)^{\frac{2}{3}}} \]
- The derivative of \( v \):
\[ \frac{dv}{dx} = \frac{1}{4}(1 - 2 \tan x)^{-\frac{3}{4}} \cdot (-2 \sec^2 x) = -\frac{1}{2}(1 - 2 \tan x)^{-\frac{3}{4}} \sec^2 x \]

### Step 4: Differentiate the denominator
The derivative of the denominator \( \sin x + \tan^2 x \):
\[ \frac{d}{dx}(\sin x + \tan^2 x) = \cos x + 2 \tan x \sec^2 x \]

### Step 5: Substitute back into the limit
Now we have:
\[ \lim_{x \to 0} \frac{\frac{2 \sin x \cos x}{3(1 + \sin^2 x)^{\frac{2}{3}}} + \frac{1}{2}(1 - 2 \tan x)^{-\frac{3}{4}} \sec^2 x}{\cos x + 2 \tan x \sec^2 x} \]

### Step 6: Evaluate the limit
Substituting \( x = 0 \):
- For the numerator:
- \( \sin x \to 0 \), \( \cos x \to 1 \), and \( \tan x \to 0 \) implies the numerator approaches \( 0 + \frac{1}{2}(1) = \frac{1}{2} \).

- For the denominator:
- \( \cos x \to 1 \) and \( \tan x \to 0 \) implies the denominator approaches \( 1 + 0 = 1 \).

Thus, the limit simplifies to:
\[ \frac{\frac{1}{2}}{1} = \frac{1}{2}. \]

### Final Answer
The limit is:

\[ \lim_{x \to 0} \frac{\sqrt[3]{1 + \sin^2 x} - \sqrt[4]{1 - 2 \tan x}}{\sin x + \tan^2 x} = \frac{1}{2}. \]