Let `f(x)=(x(3^(x)-1))/(1-cosx)" for "x ne0`. Then value of `f(0)`, which make f(x) continuous at x = 0, is

Options

A. `log3`
B. `(1)/(2) log 3`
C. `(1)/(2log3)`
D. `2log3`

Answer

To find the value of \( f(0) \) that makes the function \( f(x) = \frac{x(3^x - 1)}{1 - \cos x} \) continuous at \( x = 0 \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0.

### Step 1: Find the limit of \( f(x) \) as \( x \) approaches 0.
We start by calculating:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x(3^x - 1)}{1 - \cos x} \]

### Step 2: Simplify the numerator and denominator.
The numerator \( 3^x - 1 \) can be approximated using the exponential limit:
\[ 3^x - 1 \approx x \ln(3) \quad \text{as } x \to 0 \]
Thus, we can rewrite the numerator as:
\[ x(3^x - 1) \approx x(x \ln(3)) = x^2 \ln(3) \]

The denominator \( 1 - \cos x \) can be approximated using the small angle approximation:
\[ 1 - \cos x \approx \frac{x^2}{2} \quad \text{as } x \to 0 \]

### Step 3: Substitute the approximations into the limit.
Now substituting these approximations into the limit gives:
\[ \lim_{x \to 0} \frac{x^2 \ln(3)}{\frac{x^2}{2}} = \lim_{x \to 0} \frac{2x^2 \ln(3)}{x^2} \]

### Step 4: Cancel \( x^2 \) and evaluate the limit.
Canceling \( x^2 \) from the numerator and denominator, we get:
\[ \lim_{x \to 0} 2 \ln(3) = 2 \ln(3) \]

### Step 5: Set \( f(0) \) equal to the limit.
For \( f(x) \) to be continuous at \( x = 0 \), we need:
\[ f(0) = \lim_{x \to 0} f(x) = 2 \ln(3) \]

Thus, the value of \( f(0) \) that makes \( f(x) \) continuous at \( x = 0 \) is:
\[ \boxed{2 \ln(3)} \]