If the function `f(x) =(tan(tanx)-sin(sinx))/(tanx-sinx) (x !=0)`: is continuous at `x=0`,then find the value of `f (0)`

Answer

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

\[ f(x) = \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x} \quad (x \neq 0) \]

and ensure that the function is continuous at \( x = 0 \), we need to find the limit of \( f(x) \) as \( x \) approaches 0.

### Step-by-Step Solution:

1. **Understanding Continuity**: For \( f(x) \) to be continuous at \( x = 0 \), we need to find \( \lim_{x \to 0} f(x) \) and set \( f(0) \) equal to this limit.

2. **Finding the Limit**:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x} \]

3. **Using Taylor Series Expansions**:
- The Taylor series expansion for \( \tan x \) around \( x = 0 \) is:
\[ \tan x = x + \frac{x^3}{3} + O(x^5) \]
- The Taylor series expansion for \( \sin x \) around \( x = 0 \) is:
\[ \sin x = x - \frac{x^3}{6} + O(x^5) \]

4. **Substituting the Expansions**:
- For \( \tan(\tan x) \):
\[ \tan(\tan x) = \tan\left(x + \frac{x^3}{3} + O(x^5)\right) \approx x + \frac{x^3}{3} + O(x^5) \]
- For \( \sin(\sin x) \):
\[ \sin(\sin x) = \sin\left(x - \frac{x^3}{6} + O(x^5)\right) \approx x - \frac{x^3}{6} + O(x^5) \]

5. **Calculating the Numerator**:
\[ \tan(\tan x) - \sin(\sin x) \approx \left(x + \frac{x^3}{3}\right) - \left(x - \frac{x^3}{6}\right) = \frac{x^3}{3} + \frac{x^3}{6} = \frac{2x^3}{6} + \frac{x^3}{6} = \frac{x^3}{2} \]

6. **Calculating the Denominator**:
\[ \tan x - \sin x \approx \left(x + \frac{x^3}{3}\right) - \left(x - \frac{x^3}{6}\right) = \frac{x^3}{3} + \frac{x^3}{6} = \frac{2x^3}{6} + \frac{x^3}{6} = \frac{x^3}{2} \]

7. **Putting it Together**:
\[ f(x) = \frac{\frac{x^3}{2}}{\frac{x^3}{2}} = 1 \quad \text{for } x \neq 0 \]

8. **Finding the Limit**:
\[ \lim_{x \to 0} f(x) = 1 \]

9. **Defining \( f(0) \)**:
Since the limit exists and equals 1, we set:
\[ f(0) = 1 \]

### Final Answer:
Thus, the value of \( f(0) \) is

\[ \boxed{1} \]