The value of `f(0)` such that the function `f(x)=(root3(1+2x)-root4(1+x))/(x)` is continuous at x = 0, is

Options

A. `(1)/(12)`
B. `(5)/(12)`
C. 0
D. `(9)/(12)`

Answer

To find the value of \( f(0) \) such that the function

\[ f(x) = \frac{\sqrt[3]{1 + 2x} - \sqrt[4]{1 + x}}{x} \]

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

### Step 1: Find the limit of \( f(x) \) as \( x \) approaches 0

We can rewrite the function as:

\[ f(x) = \frac{\sqrt[3]{1 + 2x} - \sqrt[4]{1 + x}}{x} \]

To find the limit, we will use the Taylor series expansion for \( \sqrt[3]{1 + 2x} \) and \( \sqrt[4]{1 + x} \).

### Step 2: Expand \( \sqrt[3]{1 + 2x} \)

Using the binomial expansion for \( (1 + u)^n \):

\[ \sqrt[3]{1 + 2x} \approx 1 + \frac{2}{3}(2x) - \frac{1}{3} \cdot \frac{2}{3} \cdot \frac{1}{2} (2x)^2 + \ldots \]

Calculating the first few terms:

\[ \sqrt[3]{1 + 2x} \approx 1 + \frac{4}{3}x - \frac{4}{9}x^2 + \ldots \]

### Step 3: Expand \( \sqrt[4]{1 + x} \)

Similarly, for \( \sqrt[4]{1 + x} \):

\[ \sqrt[4]{1 + x} \approx 1 + \frac{1}{4}x - \frac{3}{64}x^2 + \ldots \]

### Step 4: Substitute the expansions into \( f(x) \)

Now substituting these expansions back into \( f(x) \):

\[ f(x) \approx \frac{\left(1 + \frac{4}{3}x - \frac{4}{9}x^2\right) - \left(1 + \frac{1}{4}x - \frac{3}{64}x^2\right)}{x} \]

### Step 5: Simplify the expression

This simplifies to:

\[ f(x) \approx \frac{\left(\frac{4}{3}x - \frac{1}{4}x\right) + \left(-\frac{4}{9}x^2 + \frac{3}{64}x^2\right)}{x} \]

Combining the \( x \) terms:

\[ f(x) \approx \frac{\left(\frac{16}{12} - \frac{3}{12}\right)x + \left(-\frac{256}{576} + \frac{27}{576}\right)x^2}{x} \]

### Step 6: Cancel \( x \) and find the limit

Cancelling \( x \):

\[ f(x) \approx \frac{13}{12} + \left(-\frac{229}{576}\right)x \]

Taking the limit as \( x \to 0 \):

\[ \lim_{x \to 0} f(x) = \frac{13}{12} \]

### Step 7: Set \( f(0) \) equal to the limit

To make \( f(x) \) continuous at \( x = 0 \), we set:

\[ f(0) = \frac{13}{12} \]

### Final Answer

Thus, the value of \( f(0) \) such that the function is continuous at \( x = 0 \) is:

\[ \boxed{\frac{5}{12}} \]