Prove that the function `f(x)={x/(|x|+2x^2),x!=0 and k ,x=0` remains discontinuous at `x=0`, regardless the choice of `k`

Explanation

To prove that the function

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

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

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

To find \( f(0^-) \), we consider values of \( x \) that are slightly less than 0. In this case, \( |x| = -x \) since \( x \) is negative. Thus, we can rewrite the function for \( x < 0 \):

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

Now, simplifying this expression:

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

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

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

Thus,

\[ f(0^-) = -1 \]

### Step 2: Find the right-hand limit as \( x \) approaches 0

Next, we find \( f(0^+) \) by considering values of \( x \) that are slightly greater than 0. Here, \( |x| = x \) since \( x \) is positive. Thus, we can rewrite the function for \( x > 0 \):

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

Now, simplifying this expression:

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

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

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

Thus,

\[ f(0^+) = 1 \]

### Step 3: Compare the left-hand and right-hand limits

From our calculations, we have:

\[ f(0^-) = -1 \quad \text{and} \quad f(0^+) = 1 \]

Since \( f(0^-) \neq f(0^+) \), we conclude that the limit does not exist at \( x = 0 \).

### Step 4: Conclusion about continuity

For \( f(x) \) to be continuous at \( x = 0 \), the following must hold:

1. \( f(0) \) must be defined (which it is, as \( f(0) = k \)).
2. \( \lim_{x \to 0} f(x) \) must exist.
3. \( \lim_{x \to 0} f(x) = f(0) \).

Since the left-hand limit and right-hand limit are not equal, the limit does not exist at \( x = 0 \). Therefore, the function \( f(x) \) is discontinuous at \( x = 0 \) regardless of the value of \( k \).