If f(x) ` {:( =((2^(x)-1)^(2))/(sin x.log(1+x))", if " x != 0 ),(= 2 log 2", if " x = 0 " ) :}` then , at x = 0 the function f is

Options

A. continuous at x = 0 ,1
B. discontinuous
C. not defined
D. algebraic

Answer

To determine whether the function \( f(x) \) is continuous at \( x = 0 \), we need to check if the following condition holds:

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

Given the function:

\[ f(x) = \begin{cases} \frac{(2^x - 1)^2}{\sin x \cdot \log(1+x)} & \text{if } x \neq 0 \\ 2 \log 2 & \text{if } x = 0 \end{cases} \]

### Step 1: Calculate \( f(0) \)

From the definition of the function, we have:

\[ f(0) = 2 \log 2 \]

### Step 2: Calculate \( \lim_{x \to 0} f(x) \)

We need to evaluate:

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

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

### Step 3: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can apply L'Hôpital's Rule. We differentiate the numerator and denominator:

1. **Numerator**: Differentiate \( (2^x - 1)^2 \)
- Let \( u = 2^x - 1 \), then \( \frac{du}{dx} = 2^x \ln(2) \)
- Thus, \( \frac{d}{dx}[(2^x - 1)^2] = 2(2^x - 1)(2^x \ln(2)) \)

2. **Denominator**: Differentiate \( \sin x \cdot \log(1+x) \) using the product rule:
- \( \frac{d}{dx}[\sin x] = \cos x \)
- \( \frac{d}{dx}[\log(1+x)] = \frac{1}{1+x} \)
- So, \( \frac{d}{dx}[\sin x \cdot \log(1+x)] = \sin x \cdot \frac{1}{1+x} + \log(1+x) \cdot \cos x \)

### Step 4: Evaluate the limit again

Now we evaluate:

\[ \lim_{x \to 0} \frac{2(2^x - 1)(2^x \ln(2))}{\sin x \cdot \frac{1}{1+x} + \log(1+x) \cdot \cos x} \]

Substituting \( x = 0 \):

- The numerator becomes \( 2(0)(\ln(2)) = 0 \).
- The denominator becomes \( 0 + 0 = 0 \).

We still have \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

### Step 5: Repeat L'Hôpital's Rule

Continue differentiating until we can evaluate the limit without indeterminate forms.

After applying L'Hôpital's Rule multiple times, we find:

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

### Step 6: Final Limit Calculation

Thus, we find:

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

### Step 7: Compare with \( f(0) \)

Now we compare:

\[ \lim_{x \to 0} f(x) = 0 \quad \text{and} \quad f(0) = 2 \log 2 \]

Since \( 0 \neq 2 \log 2 \), we conclude that:

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

### Conclusion

Thus, the function \( f \) is discontinuous at \( x = 0 \).