If function `f(x) = (sqrt(1+x) - root(3)(1+x))/(x)` is continuous function at x = 0, then f(0) is equal to

Options

A. 2
B. `(1)/(4)`
C. `(1)/(6)`
D. `(1)/(3)`

Answer

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

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

given that it is continuous at \( x = 0 \), we need to evaluate the limit as \( x \) approaches 0. Since directly substituting \( x = 0 \) gives us an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule.

### Step-by-step Solution:

1. **Set up the limit**:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt[3]{1+x}}{x} \]

2. **Check the form**:
Substituting \( x = 0 \):
\[ f(0) = \frac{\sqrt{1+0} - \sqrt[3]{1+0}}{0} = \frac{1 - 1}{0} = \frac{0}{0} \]
This is an indeterminate form.

3. **Apply L'Hôpital's Rule**:
Differentiate the numerator and the denominator:
- The derivative of the numerator \( \sqrt{1+x} - \sqrt[3]{1+x} \):
- For \( \sqrt{1+x} \), using the chain rule:
\[ \frac{d}{dx}(\sqrt{1+x}) = \frac{1}{2\sqrt{1+x}} \]
- For \( \sqrt[3]{1+x} \):
\[ \frac{d}{dx}(\sqrt[3]{1+x}) = \frac{1}{3(1+x)^{2/3}} \]
- Therefore, the derivative of the numerator is:
\[ \frac{1}{2\sqrt{1+x}} - \frac{1}{3(1+x)^{2/3}} \]
- The derivative of the denominator \( x \) is simply \( 1 \).

4. **Rewrite the limit**:
\[ \lim_{x \to 0} \left( \frac{\frac{1}{2\sqrt{1+x}} - \frac{1}{3(1+x)^{2/3}}}{1} \right) \]

5. **Evaluate the limit**:
Substitute \( x = 0 \):
\[ = \frac{\frac{1}{2\sqrt{1+0}} - \frac{1}{3(1+0)^{2/3}}}{1} = \frac{\frac{1}{2} - \frac{1}{3}}{1} \]
\[ = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \]

6. **Conclusion**:
Thus, we find that:
\[ f(0) = \frac{1}{6} \]

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