If `f(x)` is continuous at `x=0`, where `f(x)=(1+2x)^(1/x)", for " x!=0`, then `f(0)=`

Options

A. `e^(2)`
B. `e^(-2)`
C. `2e`
D. None of these

Answer

To solve the problem, we need to find the value of \( f(0) \) given that \( f(x) = (1 + 2x)^{\frac{1}{x}} \) for \( x \neq 0 \) and that \( f(x) \) is continuous at \( x = 0 \).

### Step-by-Step Solution:

1. **Understanding Continuity**: Since \( f(x) \) is continuous at \( x = 0 \), we have:
\[ \lim_{x \to 0} f(x) = f(0) \]
Therefore, we need to find \( \lim_{x \to 0} (1 + 2x)^{\frac{1}{x}} \).

2. **Setting Up the Limit**: We can express the limit as:
\[ L = \lim_{x \to 0} (1 + 2x)^{\frac{1}{x}} \]

3. **Using Logarithms**: To simplify the limit, we take the natural logarithm:
\[ \log L = \lim_{x \to 0} \frac{1}{x} \log(1 + 2x) \]

4. **Applying L'Hôpital's Rule**: The limit \( \lim_{x \to 0} \frac{\log(1 + 2x)}{x} \) is of the form \( \frac{0}{0} \). We can apply L'Hôpital's Rule:
\[ \log L = \lim_{x \to 0} \frac{2}{1 + 2x} \]
Evaluating this limit as \( x \to 0 \):
\[ \log L = 2 \]

5. **Exponentiating to Solve for L**: Since \( \log L = 2 \), we find \( L \) by exponentiating:
\[ L = e^2 \]

6. **Finding \( f(0) \)**: Since \( L = \lim_{x \to 0} f(x) = f(0) \), we conclude:
\[ f(0) = e^2 \]

### Final Answer:
\[ f(0) = e^2 \]