how to prove f(x)=[1-x] + [x-1] is discontinuous at x=1

To demonstrate that the function \( f(x) = [1 - x] + [x - 1] \) is discontinuous at \( x = 1 \), we need to analyze the behavior of the function as \( x \) approaches 1 from both the left and the right. Here, the brackets denote the floor function, which gives the greatest integer less than or equal to the value inside. Let's break this down step by step.

Understanding the Function

The function can be rewritten using the floor function notation:

• \( f(x) = \lfloor 1 - x \rfloor + \lfloor x - 1 \rfloor \)

Now, we will evaluate \( f(x) \) as \( x \) approaches 1 from both sides.

Evaluating from the Left: \( x \to 1^- \)

When \( x \) approaches 1 from the left (values slightly less than 1), we have:

• For \( x < 1 \), \( 1 - x \) is positive but less than 1, so \( \lfloor 1 - x \rfloor = 0 \).
• For \( x < 1 \), \( x - 1 \) is negative, so \( \lfloor x - 1 \rfloor = -1 \).

Thus, as \( x \) approaches 1 from the left:

• \( f(x) = \lfloor 1 - x \rfloor + \lfloor x - 1 \rfloor = 0 + (-1) = -1 \).

Evaluating from the Right: \( x \to 1^+ \)

Now, let's consider \( x \) approaching 1 from the right (values slightly greater than 1):

• For \( x > 1 \), \( 1 - x \) is negative, so \( \lfloor 1 - x \rfloor = -1 \).
• For \( x > 1 \), \( x - 1 \) is positive and less than 1, so \( \lfloor x - 1 \rfloor = 0 \).

Therefore, as \( x \) approaches 1 from the right:

• \( f(x) = \lfloor 1 - x \rfloor + \lfloor x - 1 \rfloor = -1 + 0 = -1 \).

Finding the Limit at \( x = 1 \)

Now, we can summarize the limits:

• \( \lim_{x \to 1^-} f(x) = -1 \)
• \( \lim_{x \to 1^+} f(x) = -1 \)

Both one-sided limits are equal, which suggests that the limit exists at \( x = 1 \). However, we also need to find the value of the function at \( x = 1 \).

Evaluating \( f(1) \)

At \( x = 1 \):

• \( f(1) = \lfloor 1 - 1 \rfloor + \lfloor 1 - 1 \rfloor = \lfloor 0 \rfloor + \lfloor 0 \rfloor = 0 + 0 = 0 \).

Conclusion on Discontinuity

Now we can compare the limit and the function value:

• \( \lim_{x \to 1} f(x) = -1 \)
• \( f(1) = 0 \)

Since the limit as \( x \) approaches 1 does not equal the value of the function at that point, we conclude that \( f(x) \) is discontinuous at \( x = 1 \). This discontinuity arises because the function jumps from -1 to 0 at that point, illustrating a classic case of discontinuity in piecewise functions.