Show `f(x) = (1)/(|x|)` has discontinuity of second kind at x = 0.

Answer

To show that the function \( f(x) = \frac{1}{|x|} \) has a discontinuity of the second kind at \( x = 0 \), we will analyze the left-hand limit (LHL) and the right-hand limit (RHL) at that point.

### Step-by-Step Solution:

1. **Define the Function**:
The function is given as:
\[ f(x) = \frac{1}{|x|} \]
We need to evaluate this function as \( x \) approaches \( 0 \) from both sides.

2. **Calculate the Left-Hand Limit (LHL)**:
The left-hand limit as \( x \) approaches \( 0 \) from the left (denoted as \( 0^- \)) is:
\[ \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{|x|} = \lim_{x \to 0^-} \frac{1}{-x} = -\infty \]
Here, since \( |x| = -x \) when \( x < 0 \), the limit approaches negative infinity.

3. **Calculate the Right-Hand Limit (RHL)**:
The right-hand limit as \( x \) approaches \( 0 \) from the right (denoted as \( 0^+ \)) is:
\[ \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1}{|x|} = \lim_{x \to 0^+} \frac{1}{x} = +\infty \]
Here, since \( |x| = x \) when \( x > 0 \), the limit approaches positive infinity.

4. **Compare the Limits**:
Now, we compare the left-hand limit and the right-hand limit:
\[ \text{LHL} = -\infty \quad \text{and} \quad \text{RHL} = +\infty \]
Since the left-hand limit and the right-hand limit are not equal, we conclude that:
\[ \text{LHL} \neq \text{RHL} \]

5. **Conclusion**:
Since the limits do not agree, the function \( f(x) \) is discontinuous at \( x = 0 \). Specifically, because the limits approach infinity in opposite directions, we classify this as a discontinuity of the second kind.

### Final Answer:
Thus, \( f(x) = \frac{1}{|x|} \) has a discontinuity of the second kind at \( x = 0 \).

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