Question

Does the function \( f(x) = \begin{cases} x+5 & \text{if } x \le 1 \\ x-5 & \text{if } x > 1 \end{cases} \) continuous at \( x = 1 \)?

Hint:

For piecewise functions, the points where the definition changes are critical points to check for continuity. The first step should always be to check if the left-hand and right-hand limits are equal at these points.

Answer 1

Step 1: Understanding the Concept:
A function \( f(x) \) is continuous at a point \( x = c \) if the following three conditions are met:
1. \( f(c) \) is defined.
2. The limit of \( f(x) \) as \( x \) approaches \( c \) exists. This means the left-hand limit (LHL) must equal the right-hand limit (RHL).
3. The limit equals the function value: \( \lim_{x \to c} f(x) = f(c) \).
Step 2: Key Formula or Approach:
We need to calculate the LHL, RHL, and \( f(1) \) and check if they are equal.
LHL: \( \lim_{x \to 1^-} f(x) \)
RHL: \( \lim_{x \to 1^+} f(x) \)
Step 3: Detailed Explanation or Calculation:
1. Calculate the Left-Hand Limit (LHL):
For \( x \to 1^- \), we use the definition \( f(x) = x + 5 \) because \( x \le 1 \).
\[ \text{LHL} = \lim_{x \to 1^-} (x + 5) = 1 + 5 = 6 \] 2. Calculate the Right-Hand Limit (RHL):
For \( x \to 1^+ \), we use the definition \( f(x) = x - 5 \) because \( x>1 \).
\[ \text{RHL} = \lim_{x \to 1^+} (x - 5) = 1 - 5 = -4 \] 3. Compare the limits:
Here, LHL = 6 and RHL = -4.
Since \( \text{LHL} \neq \text{RHL} \), the limit \( \lim_{x \to 1} f(x) \) does not exist.
Step 4: Final Answer:
Because the limit of the function as \( x \) approaches 1 does not exist, the function \( f(x) \) is not continuous (it is discontinuous) at \( x = 1 \).