For what value of k, the function
`f(x)={((x)/(|x|+2x^(2))",", x ne 0),(k",", x=0):}` is continuous at x = 0 ?

Options

A. `(1)/(2)`
B. 1
C. `(3)/(2)`
D. No value

Explanation

To determine the value of \( k \) for which the function

\[ f(x) = \begin{cases} \frac{x}{|x| + 2x^2} & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

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

### Step 1: Find the limit as \( x \) approaches 0 from the left (\( f(0^-) \))

For \( x < 0 \), we have \( |x| = -x \). Thus, the function becomes:

\[ f(x) = \frac{x}{-x + 2x^2} = \frac{x}{2x^2 - x} = \frac{x}{x(2x - 1)} = \frac{1}{2x - 1} \quad \text{(for } x < 0\text{)} \]

Now, we calculate the limit as \( x \) approaches 0 from the left:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{2x - 1} = \frac{1}{2(0) - 1} = \frac{1}{-1} = -1 \]

### Step 2: Find the limit as \( x \) approaches 0 from the right (\( f(0^+) \))

For \( x > 0 \), we have \( |x| = x \). Thus, the function becomes:

\[ f(x) = \frac{x}{x + 2x^2} = \frac{x}{x(1 + 2x)} = \frac{1}{1 + 2x} \quad \text{(for } x > 0\text{)} \]

Now, we calculate the limit as \( x \) approaches 0 from the right:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1}{1 + 2x} = \frac{1}{1 + 2(0)} = \frac{1}{1} = 1 \]

### Step 3: Set the limits equal to \( k \)

For the function to be continuous at \( x = 0 \), we need:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

This means:

\[ -1 = k \quad \text{and} \quad 1 = k \]

Since \( -1 \neq 1 \), there is no single value of \( k \) that can satisfy both conditions. Therefore, the function cannot be made continuous at \( x = 0 \) for any value of \( k \).

### Conclusion

The function \( f(x) \) is not continuous at \( x = 0 \) for any value of \( k \).

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