The value of f(0) so that the function `f(x) = (log(1+x^(2) tanx))/(sin x^(3)), (x ne 0)` ` is continuous at x = 0 is:

Options

A. 1
B. 2
C. 3
D. none

Answer

To find the value of \( f(0) \) such that the function

\[ f(x) = \frac{\log(1 + x^2 \tan x)}{\sin x^3}, \quad (x \neq 0) \]

is continuous at \( x = 0 \), we need to ensure that

\[ \lim_{x \to 0} f(x) = f(0). \]

### Step 1: Calculate the limit as \( x \to 0 \)

We start by calculating the limit:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\log(1 + x^2 \tan x)}{\sin x^3}. \]

### Step 2: Analyze the limit

As \( x \to 0 \), both the numerator and denominator approach 0, creating an indeterminate form \( \frac{0}{0} \). Thus, we can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule

Using L'Hôpital's Rule, we differentiate the numerator and denominator:

1. **Differentiate the numerator**:
\[ \frac{d}{dx} \log(1 + x^2 \tan x) = \frac{1}{1 + x^2 \tan x} \cdot \frac{d}{dx}(x^2 \tan x). \]
Using the product rule:
\[ \frac{d}{dx}(x^2 \tan x) = 2x \tan x + x^2 \sec^2 x. \]

2. **Differentiate the denominator**:
\[ \frac{d}{dx}(\sin x^3) = 3x^2 \cos x^3. \]

### Step 4: Rewrite the limit

Now, we rewrite the limit using these derivatives:

\[ \lim_{x \to 0} \frac{\frac{1}{1 + x^2 \tan x} \cdot (2x \tan x + x^2 \sec^2 x)}{3x^2 \cos x^3}. \]

### Step 5: Simplify the limit

As \( x \to 0 \):

- \( \tan x \approx x \) and \( \sec^2 x \approx 1 \).
- Thus, \( x^2 \tan x \approx x^3 \) and \( 1 + x^2 \tan x \approx 1 \).

Substituting these approximations gives:

\[ \lim_{x \to 0} \frac{2x^2 + x^2}{3x^2} = \lim_{x \to 0} \frac{3x^2}{3x^2} = 1. \]

### Step 6: Set \( f(0) \)

To ensure continuity at \( x = 0 \), we set:

\[ f(0) = \lim_{x \to 0} f(x) = 1. \]

### Final Answer

Thus, the value of \( f(0) \) that makes the function continuous at \( x = 0 \) is:

\[ \boxed{1}. \]