If `f(x)` is continuous at `x=0`, where `f(x)=((3^(sin x)-1)^(2))/(x log (1-x))`, for `x!=0`, then `f(0)=`

Options

A. `(log 3)^(2)`
B. log 9
C. `1/2 ` log 3
D. log 3

Answer

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

\[ f(x) = \frac{(3^{\sin x} - 1)^2}{x \log(1 - x)} \]

given that \( f(x) \) is continuous at \( x = 0 \), we need to evaluate the limit as \( x \) approaches 0:

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

### Step 1: Evaluate the limit

First, we substitute \( x = 0 \) into the function:

\[ f(0) = \frac{(3^{\sin(0)} - 1)^2}{0 \cdot \log(1 - 0)} \]

Since \( \sin(0) = 0 \), we have:

\[ 3^{\sin(0)} - 1 = 3^0 - 1 = 1 - 1 = 0 \]

Thus, the limit takes the form \( \frac{0^2}{0} \), which is indeterminate. We need to apply L'Hôpital's rule or find another way to evaluate the limit.

### Step 2: Rewrite the expression

Using the property of limits, we can rewrite the numerator:

\[ 3^{\sin x} - 1 \approx \sin x \cdot \log 3 \quad \text{as } x \to 0 \]

This is derived from the fact that:

\[ \lim_{x \to 0} \frac{3^{\sin x} - 1}{\sin x} = \log 3 \]

Thus, we can express the numerator as:

\[ (3^{\sin x} - 1)^2 \approx (\sin x \cdot \log 3)^2 = \sin^2 x \cdot (\log 3)^2 \]

### Step 3: Substitute back into the limit

Now substitute this back into the limit:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin^2 x \cdot (\log 3)^2}{x \log(1 - x)} \]

### Step 4: Simplify the denominator

Next, we know that:

\[ \log(1 - x) \approx -x \quad \text{as } x \to 0 \]

Thus, we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{\sin^2 x \cdot (\log 3)^2}{x \cdot (-x)} = \lim_{x \to 0} \frac{\sin^2 x \cdot (\log 3)^2}{-x^2} \]

### Step 5: Apply the limit

Using the fact that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):

\[ \lim_{x \to 0} \frac{\sin^2 x}{x^2} = 1 \]

Thus, we have:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin^2 x \cdot (\log 3)^2}{-x^2} = -(\log 3)^2 \]

### Conclusion

Therefore, we find that:

\[ f(0) = -(\log 3)^2 \]