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 evaluate the left-hand limit (LHL) and the right-hand limit (RHL) as \( x \) approaches 0.

### Step 1: Find the Left-Hand Limit (LHL)

We need to calculate the limit of \( f(x) \) as \( x \) approaches 0 from the left (i.e., \( x \to 0^- \)).

For \( x < 0 \), we have:

\[ f(x) = \frac{x}{|x| + 2x^2} = \frac{x}{-x + 2x^2} = \frac{x}{-x + 2x^2} = \frac{x}{-x(1 - 2x)} = \frac{1}{-1 + 2x} \]

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

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

### Step 2: Find the Right-Hand Limit (RHL)

Now, we calculate the limit of \( f(x) \) as \( x \) approaches 0 from the right (i.e., \( x \to 0^+ \)).

For \( x > 0 \), we have:

\[ f(x) = \frac{x}{|x| + 2x^2} = \frac{x}{x + 2x^2} = \frac{x}{x(1 + 2x)} = \frac{1}{1 + 2x} \]

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

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

### Step 3: Compare the Limits and the Function Value at \( x = 0 \)

We found:

- Left-hand limit: \( \lim_{x \to 0^-} f(x) = -1 \)
- Right-hand limit: \( \lim_{x \to 0^+} f(x) = 1 \)

Since the left-hand limit does not equal the right-hand limit:

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

### Conclusion

Since the left-hand limit and right-hand limit are not equal, we conclude that \( f(x) \) is discontinuous at \( x = 0 \) regardless of the value of \( k \).