Let ` f(x) = (x(2^(x)-1))/( 1- cos x)` for ` x ne 0` what choice of f(0) , if any, will make f (x) continuous at x = 0 ?

Options

A. log 2
B. ` 1/2 log 2`
C. ` `1/( 2 log 2)`
D. 2 log 2

Answer

To determine the value of \( f(0) \) that will make the function \( f(x) = \frac{x(2^x - 1)}{1 - \cos x} \) continuous at \( x = 0 \), we need to find the limit of \( f(x) \) as \( x \) approaches 0 and set \( f(0) \) equal to that limit.

### Step-by-Step Solution:

1. **Find the limit as \( x \) approaches 0**:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x(2^x - 1)}{1 - \cos x} \]

2. **Evaluate the numerator**:
- As \( x \to 0 \), \( 2^x - 1 \) can be approximated using the limit property:
\[ \lim_{x \to 0} \frac{2^x - 1}{x} = \ln(2) \]
Therefore, we can write:
\[ 2^x - 1 \approx x \ln(2) \text{ as } x \to 0 \]
Thus, the numerator becomes:
\[ x(2^x - 1) \approx x \cdot x \ln(2) = x^2 \ln(2) \]

3. **Evaluate the denominator**:
- Using the limit property for \( 1 - \cos x \):
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \]
Therefore, we can write:
\[ 1 - \cos x \approx \frac{x^2}{2} \text{ as } x \to 0 \]

4. **Substituting the approximations into the limit**:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x^2 \ln(2)}{\frac{x^2}{2}} = \lim_{x \to 0} \frac{2 \ln(2) x^2}{x^2} = 2 \ln(2) \]

5. **Setting \( f(0) \)**:
To make \( f(x) \) continuous at \( x = 0 \), we set:
\[ f(0) = 2 \ln(2) \]

### Conclusion:
Thus, the choice of \( f(0) \) that will make \( f(x) \) continuous at \( x = 0 \) is:
\[ \boxed{2 \ln(2)} \]