Question

Test whether the function \(f: \mathbb{R} \to \mathbb{R}\) defined by \[ f(x) = \begin{cases} x+5, & \text{if } x \le 1 \\ x-5, & \text{if } x > 1 \end{cases} \] is continuous at \(x=1\).

Hint:

For piecewise functions, the points where the definition changes are the most likely points of discontinuity. Always check the continuity at these points by comparing the left-hand limit, right-hand limit, and the function's value.

Answer 1

Step 1: Understanding the Concept:
A function \(f(x)\) is continuous at a point \(x=c\) if:
1. \(f(c)\) is defined.
2. \(\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)\) (the limit exists).
3. \(\lim_{x \to c} f(x) = f(c)\).

Step 2: Key Formula or Approach:
At \(x=1\), compute:
1. LHL: \(\lim_{x \to 1^-} f(x)\)
2. RHL: \(\lim_{x \to 1^+} f(x)\)
3. \(f(1)\)

Step 3: Detailed Calculation:
The function is defined as:
\[ f(x) = \begin{cases} x+5, & x \le 1 \\ x-5, & x > 1 \end{cases} \]

1. Left-Hand Limit (LHL):
\(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+5) = 6\).

2. Right-Hand Limit (RHL):
\(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x-5) = -4\).

3. Value of the function:
\(f(1) = 1+5 = 6\).

Since LHL = 6 and RHL = -4, they are not equal. Hence, \(\lim_{x \to 1} f(x)\) does not exist.

Step 4: Final Answer:
The function \(f(x)\) is discontinuous at \(x=1\).