Find the value of f(0) so that the function `f(x)=([log (1+x//12)-log (1-x//8)])/x , x ne 0` is continuous on [0,8].

Answer

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

\[ f(x) = \frac{\log(1 + \frac{x}{12}) - \log(1 - \frac{x}{8})}{x}, \quad x \neq 0 \]

is continuous on the interval \([0, 8]\), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0.

### Step-by-step Solution:

1. **Understanding Continuity**:
A function is continuous at a point if the limit of the function as it approaches that point equals the function's value at that point. Therefore, we need to find:

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

2. **Substituting into the Limit**:
We substitute \( f(x) \) into the limit:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\log(1 + \frac{x}{12}) - \log(1 - \frac{x}{8})}{x} \]

As \( x \to 0 \), both the numerator and denominator approach 0, resulting in the indeterminate form \( \frac{0}{0} \).

3. **Applying L'Hôpital's Rule**:
Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

where \( f(x) \) is the numerator and \( g(x) \) is the denominator.

4. **Differentiating the Numerator and Denominator**:
- Differentiate the numerator:

\[ \frac{d}{dx} \left( \log(1 + \frac{x}{12}) - \log(1 - \frac{x}{8}) \right) = \frac{1}{1 + \frac{x}{12}} \cdot \frac{1}{12} - \left( -\frac{1}{1 - \frac{x}{8}} \cdot \frac{1}{8} \right) \]

Simplifying this gives:

\[ \frac{1}{12(1 + \frac{x}{12})} + \frac{1}{8(1 - \frac{x}{8})} \]

- Differentiate the denominator \( x \):

\[ \frac{d}{dx}(x) = 1 \]

5. **Evaluating the Limit**:
Now we can rewrite the limit using L'Hôpital's Rule:

\[ \lim_{x \to 0} \left( \frac{\frac{1}{12(1 + \frac{x}{12})} + \frac{1}{8(1 - \frac{x}{8})}}{1} \right) \]

Substituting \( x = 0 \):

\[ = \frac{1}{12(1 + 0)} + \frac{1}{8(1 - 0)} = \frac{1}{12} + \frac{1}{8} \]

To add these fractions, find a common denominator (which is 24):

\[ \frac{1}{12} = \frac{2}{24}, \quad \frac{1}{8} = \frac{3}{24} \]

Therefore:

\[ \frac{1}{12} + \frac{1}{8} = \frac{2}{24} + \frac{3}{24} = \frac{5}{24} \]

6. **Final Value of \( f(0) \)**:
Thus, we find:

\[ f(0) = \frac{5}{24} \]

### Conclusion:
The value of \( f(0) \) so that the function is continuous on \([0, 8]\) is

\[ \boxed{\frac{5}{24}} \]