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 ensure that

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

### Step 1: Evaluate the limit as \( x \) approaches 0

We start by calculating the limit:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sqrt[3]{1 + 2x} - \sqrt[4]{1 + x}}{x}. \]

### Step 2: Apply Taylor series expansions

We can use Taylor series expansions for \( \sqrt[3]{1 + 2x} \) and \( \sqrt[4]{1 + x} \) around \( x = 0 \).

1. For \( \sqrt[3]{1 + 2x} \):

\[ \sqrt[3]{1 + 2x} \approx 1 + \frac{2}{3} \cdot 2x - \frac{1}{3} \cdot \frac{4}{9} \cdot (2x)^2 + \ldots = 1 + \frac{2}{3} \cdot 2x - \frac{8}{27} x^2 + \ldots \]

2. For \( \sqrt[4]{1 + x} \):

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

### Step 3: Substitute the expansions into the limit

Substituting these expansions into our limit gives:

\[ \lim_{x \to 0} \frac{\left(1 + \frac{4}{3} x - \frac{8}{27} x^2\right) - \left(1 + \frac{1}{4} x - \frac{3}{64} x^2\right)}{x}. \]

### Step 4: Simplify the expression

Now simplify the numerator:

\[ \frac{4}{3} x - \frac{1}{4} x - \left(-\frac{8}{27} x^2 + \frac{3}{64} x^2\right). \]

Combining the linear terms:

\[ \left(\frac{4}{3} - \frac{1}{4}\right)x = \left(\frac{16}{12} - \frac{3}{12}\right)x = \frac{13}{12} x. \]

Combining the quadratic terms:

\[ -\left(-\frac{8}{27} + \frac{3}{64}\right)x^2. \]

### Step 5: Calculate the limit

Now, substituting back into the limit:

\[ \lim_{x \to 0} \frac{\frac{13}{12} x + \text{higher order terms}}{x} = \frac{13}{12} + \text{higher order terms vanish as } x \to 0. \]

Thus, we have:

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

### Step 6: Set \( f(0) \) equal to the limit

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

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

### Final Answer

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

\[ \boxed{\frac{13}{12}}. \]