If the function `f(x)=(2x-sin^(-1)x)/(2x+tan^(-1)x)` is continuous at each point in its domain, then what is the value of f(0)?

Options

A. `-(1)/(3)`
B. `(1)/(3)`
C. `(2)/(3)`
D. 2

Answer

To find the value of \( f(0) \) for the function

\[ f(x) = \frac{2x - \sin^{-1}(x)}{2x + \tan^{-1}(x)}, \]

we need to ensure that the function is continuous at \( x = 0 \). Since direct substitution leads to an indeterminate form \( \frac{0}{0} \), we will use the limit approach.

### Step-by-Step Solution:

1. **Identify the limit to evaluate:**
We need to find:
\[ f(0) = \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{2x - \sin^{-1}(x)}{2x + \tan^{-1}(x)}. \]

2. **Divide numerator and denominator by \( x \):**
To simplify the limit, we divide both the numerator and the denominator by \( x \):
\[ \lim_{x \to 0} \frac{2 - \frac{\sin^{-1}(x)}{x}}{2 + \frac{\tan^{-1}(x)}{x}}. \]

3. **Evaluate the limits of \( \frac{\sin^{-1}(x)}{x} \) and \( \frac{\tan^{-1}(x)}{x} \):**
We know from standard limits that:
\[ \lim_{x \to 0} \frac{\sin^{-1}(x)}{x} = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{\tan^{-1}(x)}{x} = 1. \]

4. **Substitute the limits into the expression:**
Now substituting these limits into our expression gives:
\[ \lim_{x \to 0} \frac{2 - 1}{2 + 1} = \frac{1}{3}. \]

5. **Conclusion:**
Therefore, the value of \( f(0) \) is:
\[ f(0) = \frac{1}{3}. \]

### Final Answer:
The value of \( f(0) \) is \( \frac{1}{3} \).