Prove that the function f defined by `f(x) = {{:((x)/(|x|+2x^(2)), if x ne 0),(k, if x = 0):}` remains discontinuous at `x = 0`, regardings the choice iof k.

Explanation

To prove that the function \( f \) defined by

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

remains discontinuous at \( x = 0 \) regardless of the choice of \( k \), we will follow these steps:

### Step 1: Find \( f(0) \)

From the definition of the function, we have:

\[ f(0) = k \]

### Step 2: Find the right-hand limit \( f(0^+) \)

To find \( f(0^+) \), we consider the limit as \( x \) approaches 0 from the right (i.e., \( x \to 0^+ \)):

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

Since \( x \) is positive when approaching from the right, \( |x| = x \). Thus, we have:

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

Factoring out \( x \) from the denominator:

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

As \( x \to 0^+ \), \( 2x \to 0 \), so:

\[ f(0^+) = \frac{1}{1 + 0} = 1 \]

### Step 3: Find the left-hand limit \( f(0^-) \)

Now, we find \( f(0^-) \) by taking the limit as \( x \) approaches 0 from the left (i.e., \( x \to 0^- \)):

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

Since \( x \) is negative when approaching from the left, \( |x| = -x \). Thus, we have:

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

Factoring out \( x \) from the denominator:

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

As \( x \to 0^- \), \( 2x \to 0 \), so:

\[ f(0^-) = \frac{1}{-1 + 0} = -1 \]

### Step 4: Compare limits and value at \( x = 0 \)

Now we compare the limits and the value at \( x = 0 \):

- \( f(0) = k \)
- \( f(0^+) = 1 \)
- \( f(0^-) = -1 \)

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

\[ f(0^+) = f(0) = f(0^-) \]

This means we need \( k = 1 \) and \( k = -1 \). However, since \( 1 \neq -1 \), we conclude that the left-hand limit and the right-hand limit are not equal. Therefore, the function cannot be continuous at \( x = 0 \) for any value of \( k \).

### Conclusion

Thus, we have shown that the function \( f \) remains discontinuous at \( x = 0 \) regardless of the choice of \( k \).

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