If `f(x)` is continuous at `x=0`, where `f(x)=(sqrt(1+x)-root3(1+x))/(x)`, for `x!=0`, then `f(0)=`

Options

A. `1/3`
B. `1/2`
C. `5/6`
D. `1/6`

Answer

To find the value of \( f(0) \) for the function defined as

\[ f(x) = \frac{\sqrt{1+x} - \sqrt[3]{1+x}}{x} \quad \text{for } x \neq 0, \]

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

### Step-by-Step Solution:

1. **Identify the limit**: We need to find

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt[3]{1+x}}{x}. \]

2. **Substituting \( x = 0 \)**: If we substitute \( x = 0 \) directly into the function, we get:

\[ f(0) = \frac{\sqrt{1+0} - \sqrt[3]{1+0}}{0} = \frac{1 - 1}{0} = \frac{0}{0}, \]

which is an indeterminate form. Therefore, we can apply L'Hôpital's Rule.

3. **Apply L'Hôpital's Rule**: According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:

- **Differentiate the numerator**:
- The derivative of \( \sqrt{1+x} \) is \( \frac{1}{2\sqrt{1+x}} \).
- The derivative of \( \sqrt[3]{1+x} \) is \( \frac{1}{3(1+x)^{2/3}} \).
- Thus, the derivative of the numerator is:

\[ \frac{1}{2\sqrt{1+x}} - \frac{1}{3(1+x)^{2/3}}. \]

- **Differentiate the denominator**: The derivative of \( x \) is \( 1 \).

4. **Rewrite the limit**: Now, we can rewrite the limit as:

\[ \lim_{x \to 0} \left( \frac{\frac{1}{2\sqrt{1+x}} - \frac{1}{3(1+x)^{2/3}}}{1} \right). \]

5. **Evaluate the limit**: Now substitute \( x = 0 \):

\[ \lim_{x \to 0} \left( \frac{\frac{1}{2\sqrt{1+0}} - \frac{1}{3(1+0)^{2/3}}}{1} \right) = \frac{\frac{1}{2 \cdot 1} - \frac{1}{3 \cdot 1}}{1} = \frac{\frac{1}{2} - \frac{1}{3}}{1}. \]

6. **Simplify the expression**: To simplify \( \frac{1}{2} - \frac{1}{3} \):

\[ \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}. \]

7. **Conclusion**: Therefore, we find that

\[ \lim_{x \to 0} f(x) = \frac{1}{6}. \]

Since \( f(x) \) is continuous at \( x = 0 \), we conclude that

\[ f(0) = \frac{1}{6}. \]

### Final Answer:
\[ f(0) = \frac{1}{6}. \]